Given a positive integer N, the task is to count the total number of set bits in binary representation of all numbers from 1 to N.
Examples:
Input: N = 3
Output: 4
Explanation: Numbers from 1 to 3: {1, 2, 3}
Binary Representation of 1: 01 -> Set bits = 1
Binary Representation of 2: 10 -> Set bits = 1
Binary Representation of 3: 11 -> Set bits = 2
Total set bits from 1 till 3 = 1 + 1 + 2 = 4Input: N = 6
Output: 9Input: N = 7
Output: 12
Source: Amazon Interview Question
Count total set bits by converting each number into its Binary Representation:
The idea is to convert each number from 1 till N into binary, and count the set bits in each number separately.
Follow the steps below to understand how:
- Traverse a loop from 1 to N
- For each integer in 1 to N:
- Convert the number to its binary representation
- Add the count of 1s in the binary representation to the answer.
- Return the total set bits count.
Below is the implementation of the above approach:
C++
// A simple program to count set bits// in all numbers from 1 to n.#include <iostream>using namespace std;// A utility function to count set bits// in a number xunsigned int countSetBitsUtil(unsigned int x);// Returns count of set bits present in all// numbers from 1 to nunsigned int countSetBits(unsigned int n){ int bitCount = 0; // initialize the result for (int i = 1; i <= n; i++) bitCount += countSetBitsUtil(i); return bitCount;}// A utility function to count set bits// in a number xunsigned int countSetBitsUtil(unsigned int x){ if (x <= 0) return 0; return (x % 2 == 0 ? 0 : 1) + countSetBitsUtil(x / 2);}// Driver program to test above functionsint main(){ int n = 4; cout << "Total set bit count is " << countSetBits(n); return 0;}// This code is contributed by shivanisinghss2110. |
C
// A simple program to count set bits// in all numbers from 1 to n.#include <stdio.h>// A utility function to count set bits// in a number xunsigned int countSetBitsUtil(unsigned int x);// Returns count of set bits present in all// numbers from 1 to nunsigned int countSetBits(unsigned int n){ int bitCount = 0; // initialize the result for (int i = 1; i <= n; i++) bitCount += countSetBitsUtil(i); return bitCount;}// A utility function to count set bits// in a number xunsigned int countSetBitsUtil(unsigned int x){ if (x <= 0) return 0; return (x % 2 == 0 ? 0 : 1) + countSetBitsUtil(x / 2);}// Driver program to test above functionsint main(){ int n = 4; printf("Total set bit count is %d", countSetBits(n)); return 0;} |
Java
// A simple program to count set bits// in all numbers from 1 to n.class GFG { // Returns count of set bits present // in all numbers from 1 to n static int countSetBits(int n) { // initialize the result int bitCount = 0; for (int i = 1; i <= n; i++) bitCount += countSetBitsUtil(i); return bitCount; } // A utility function to count set bits // in a number x static int countSetBitsUtil(int x) { if (x <= 0) return 0; return (x % 2 == 0 ? 0 : 1) + countSetBitsUtil(x / 2); } // Driver program public static void main(String[] args) { int n = 4; System.out.print("Total set bit count is "); System.out.print(countSetBits(n)); }}// This code is contributed by// Smitha Dinesh Semwal |
Python3
# A simple program to count set bits# in all numbers from 1 to n.# Returns count of set bits present in all# numbers from 1 to ndef countSetBits(n): # initialize the result bitCount = 0 for i in range(1, n + 1): bitCount += countSetBitsUtil(i) return bitCount# A utility function to count set bits# in a number xdef countSetBitsUtil(x): if (x <= 0): return 0 return (0 if int(x % 2) == 0 else 1) + countSetBitsUtil(int(x / 2))# Driver programif __name__ == '__main__': n = 4 print("Total set bit count is", countSetBits(n))# This code is contributed by# Smitha Dinesh Semwal |
C#
// A simple C# program to count set bits// in all numbers from 1 to n.using System;class GFG { // Returns count of set bits present // in all numbers from 1 to n static int countSetBits(int n) { // initialize the result int bitCount = 0; for (int i = 1; i <= n; i++) bitCount += countSetBitsUtil(i); return bitCount; } // A utility function to count set bits // in a number x static int countSetBitsUtil(int x) { if (x <= 0) return 0; return (x % 2 == 0 ? 0 : 1) + countSetBitsUtil(x / 2); } // Driver program public static void Main() { int n = 4; Console.Write("Total set bit count is "); Console.Write(countSetBits(n)); }}// This code is contributed by Sam007 |
Javascript
// A simple program to count set bits// in all numbers from 1 to n. // Returns count of set bits present // in all numbers from 1 to n function countSetBits(n) { // initialize the result let bitCount = 0; for (let i = 1; i <= n; i++) { bitCount += countSetBitsUtil(i); } return bitCount; } // A utility function to count set bits // in a number x function countSetBitsUtil(x) { if (x <= 0) { return 0; } return (x % 2 == 0 ? 0 : 1) + countSetBitsUtil(Math.floor(x/2)); } // Driver program let n = 4; console.log("Total set bit count is "); console.log(countSetBits(n)); // This code is contributed by rag2127 |
PHP
<?php // A simple program to count set bits// in all numbers from 1 to n.// Returns count of set bits present // in all numbers from 1 to nfunction countSetBits($n){ $bitCount = 0; // initialize the result for ($i = 1; $i <= $n; $i++) $bitCount += countSetBitsUtil($i); return $bitCount;}// A utility function to count // set bits in a number xfunction countSetBitsUtil($x){ if ($x <= 0) return 0; return ($x % 2 == 0 ? 0 : 1) + countSetBitsUtil($x / 2);}// Driver Code$n = 4;echo "Total set bit count is " . countSetBits($n);// This code is contributed by ChitraNayal?> |
Output
Total set bit count is 5
Time Complexity: O(N*log(N)), where N is the given integer and log(N) time is used for the binary conversion of the number.
Auxiliary Space: O(1).
Method 2 (Simple and efficient than Method 1)
If we observe bits from rightmost side at distance i than bits get inverted after 2^i position in vertical sequence.
for example n = 5;
0 = 0000
1 = 0001
2 = 0010
3 = 0011
4 = 0100
5 = 0101
Observe the right most bit (i = 0) the bits get flipped after (2^0 = 1)
Observe the 3rd rightmost bit (i = 2) the bits get flipped after (2^2 = 4)
So, We can count bits in vertical fashion such that at i’th right most position bits will be get flipped after 2^i iteration;
C++
#include <bits/stdc++.h>using namespace std;// Function which counts set bits from 0 to nint countSetBits(int n){ int i = 0; // ans store sum of set bits from 0 to n int ans = 0; // while n greater than equal to 2^i while ((1 << i) <= n) { // This k will get flipped after // 2^i iterations bool k = 0; // change is iterator from 2^i to 1 int change = 1 << i; // This will loop from 0 to n for // every bit position for (int j = 0; j <= n; j++) { ans += k; if (change == 1) { k = !k; // When change = 1 flip the bit change = 1 << i; // again set change to 2^i } else { change--; } } // increment the position i++; } return ans;}// Main Functionint main(){ int n = 17; cout << countSetBits(n) << endl; return 0;} |
Java
public class GFG { // Function which counts set bits from 0 to n static int countSetBits(int n) { int i = 0; // ans store sum of set bits from 0 to n int ans = 0; // while n greater than equal to 2^i while ((1 << i) <= n) { // This k will get flipped after // 2^i iterations boolean k = false; // change is iterator from 2^i to 1 int change = 1 << i; // This will loop from 0 to n for // every bit position for (int j = 0; j <= n; j++) { if (k == true) ans += 1; else ans += 0; if (change == 1) { // When change = 1 flip the bit k = !k; // again set change to 2^i change = 1 << i; } else { change--; } } // increment the position i++; } return ans; } // Driver program public static void main(String[] args) { int n = 17; System.out.println(countSetBits(n)); }}// This code is contributed by Sam007. |
Python3
# Function which counts set bits from 0 to ndef countSetBits(n): i = 0 # ans store sum of set bits from 0 to n ans = 0 # while n greater than equal to 2^i while ((1 << i) <= n): # This k will get flipped after # 2^i iterations k = 0 # change is iterator from 2^i to 1 change = 1 << i # This will loop from 0 to n for # every bit position for j in range(0, n+1): ans += k if change == 1: # When change = 1 flip the bit k = not k # again set change to 2^i change = 1 << i else: change -= 1 # increment the position i += 1 return ans# Driver codeif __name__ == "__main__": n = 17 print(countSetBits(n))# This code is contributed by ANKITRAI1 |
C#
using System;class GFG { static int countSetBits(int n) { int i = 0; // ans store sum of set bits from 0 to n int ans = 0; // while n greater than equal to 2^i while ((1 << i) <= n) { // This k will get flipped after // 2^i iterations bool k = false; // change is iterator from 2^i to 1 int change = 1 << i; // This will loop from 0 to n for // every bit position for (int j = 0; j <= n; j++) { if (k == true) ans += 1; else ans += 0; if (change == 1) { // When change = 1 flip the bit k = !k; // again set change to 2^i change = 1 << i; } else { change--; } } // increment the position i++; } return ans; } // Driver program static void Main() { int n = 17; Console.Write(countSetBits(n)); }}// This code is contributed by Sam007 |
Javascript
<script>// Javascript program for the above approach// Function which counts set bits from 0 to nfunction countSetBits(n){ let i = 0; // ans store sum of set bits from 0 to n let ans = 0; // while n greater than equal to 2^i while ((1 << i) <= n) { // This k will get flipped after // 2^i iterations let k = 0; // change is iterator from 2^i to 1 let change = 1 << i; // This will loop from 0 to n for // every bit position for (let j = 0; j <= n; j++) { ans += k; if (change == 1) { // When change = 1 flip the bit k = !k; // again set change to 2^i change = 1 << i; } else { change--; } } // increment the position i++; } return ans;} // Driver Code let n = 17; document.write(countSetBits(n));</script> |
PHP
<?php// Function which counts // set bits from 0 to nfunction countSetBits($n){ $i = 0; // ans store sum of set // bits from 0 to n $ans = 0; // while n greater than // equal to 2^i while ((1 << $i) <= $n) { // This k will get flipped // after 2^i iterations $k = 0; // change is iterator // from 2^i to 1 $change = 1 << $i; // This will loop from 0 to n // for every bit position for ($j = 0; $j <= $n; $j++) { $ans += $k; if ($change == 1) { // When change = 1 flip // the bit $k = !$k; // again set change to 2^i $change = 1 << $i; } else { $change--; } } // increment the position $i++; } return $ans;}// Driver code$n = 17;echo(countSetBits($n));// This code is contributed by Smitha?> |
Output
35
Time Complexity: O(k*n)
where k = number of bits to represent number n
k <= 64
Method 3 (Tricky)
If the input number is of the form 2^b -1 e.g., 1, 3, 7, 15.. etc, the number of set bits is b * 2^(b-1). This is because for all the numbers 0 to (2^b)-1, if you complement and flip the list you end up with the same list (half the bits are on, half off).
If the number does not have all set bits, then some position m is the position of leftmost set bit. The number of set bits in that position is n – (1 << m) + 1. The remaining set bits are in two parts:
1) The bits in the (m-1) positions down to the point where the leftmost bit becomes 0, and
2) The 2^(m-1) numbers below that point, which is the closed form above.
An easy way to look at it is to consider the number 6:
0|0 0
0|0 1
0|1 0
0|1 1
-|--
1|0 0
1|0 1
1|1 0
The leftmost set bit is in position 2 (positions are considered starting from 0). If we mask that off what remains is 2 (the “1 0” in the right part of the last row.) So the number of bits in the 2nd position (the lower left box) is 3 (that is, 2 + 1). The set bits from 0-3 (the upper right box above) is 2*2^(2-1) = 4. The box in the lower right is the remaining bits we haven’t yet counted, and is the number of set bits for all the numbers up to 2 (the value of the last entry in the lower right box) which can be figured recursively.
C++
#include <bits/stdc++.h>// A O(Logn) complexity program to count// set bits in all numbers from 1 to nusing namespace std;/* Returns position of leftmost set bit.The rightmost position is consideredas 0 */unsigned int getLeftmostBit(int n){ int m = 0; while (n > 1) { n = n >> 1; m++; } return m;}/* Given the position of previous leftmostset bit in n (or an upper bound onleftmost position) returns the newposition of leftmost set bit in n */unsigned int getNextLeftmostBit(int n, int m){ unsigned int temp = 1 << m; while (n < temp) { temp = temp >> 1; m--; } return m;}// The main recursive function used by countSetBits()unsigned int _countSetBits(unsigned int n, int m);// Returns count of set bits present in// all numbers from 1 to nunsigned int countSetBits(unsigned int n){ // Get the position of leftmost set // bit in n. This will be used as an // upper bound for next set bit function int m = getLeftmostBit(n); // Use the position return _countSetBits(n, m);}unsigned int _countSetBits(unsigned int n, int m){ // Base Case: if n is 0, then set bit // count is 0 if (n == 0) return 0; /* get position of next leftmost set bit */ m = getNextLeftmostBit(n, m); // If n is of the form 2^x-1, i.e., if n // is like 1, 3, 7, 15, 31, .. etc, // then we are done. // Since positions are considered starting // from 0, 1 is added to m if (n == ((unsigned int)1 << (m + 1)) - 1) return (unsigned int)(m + 1) * (1 << m); // update n for next recursive call n = n - (1 << m); return (n + 1) + countSetBits(n) + m * (1 << (m - 1));}// Driver codeint main(){ int n = 17; cout << "Total set bit count is " << countSetBits(n); return 0;}// This code is contributed by rathbhupendra |
C
// A O(Logn) complexity program to count// set bits in all numbers from 1 to n#include <stdio.h>/* Returns position of leftmost set bit. The rightmost position is considered as 0 */unsigned int getLeftmostBit(int n){ int m = 0; while (n > 1) { n = n >> 1; m++; } return m;}/* Given the position of previous leftmost set bit in n (or an upper bound on leftmost position) returns the new position of leftmost set bit in n */unsigned int getNextLeftmostBit(int n, int m){ unsigned int temp = 1 << m; while (n < temp) { temp = temp >> 1; m--; } return m;}// The main recursive function used by countSetBits()unsigned int _countSetBits(unsigned int n, int m);// Returns count of set bits present in// all numbers from 1 to nunsigned int countSetBits(unsigned int n){ // Get the position of leftmost set // bit in n. This will be used as an // upper bound for next set bit function int m = getLeftmostBit(n); // Use the position return _countSetBits(n, m);}unsigned int _countSetBits(unsigned int n, int m){ // Base Case: if n is 0, then set bit // count is 0 if (n == 0) return 0; /* get position of next leftmost set bit */ m = getNextLeftmostBit(n, m); // If n is of the form 2^x-1, i.e., if n // is like 1, 3, 7, 15, 31, .. etc, // then we are done. // Since positions are considered starting // from 0, 1 is added to m if (n == ((unsigned int)1 << (m + 1)) - 1) return (unsigned int)(m + 1) * (1 << m); // update n for next recursive call n = n - (1 << m); return (n + 1) + countSetBits(n) + m * (1 << (m - 1));}// Driver program to test above functionsint main(){ int n = 17; printf("Total set bit count is %d", countSetBits(n)); return 0;} |
Java
// Java A O(Logn) complexity program to count// set bits in all numbers from 1 to nimport java.io.*;class GFG { /* Returns position of leftmost set bit. The rightmost position is considered as 0 */ static int getLeftmostBit(int n) { int m = 0; while (n > 1) { n = n >> 1; m++; } return m; } /* Given the position of previous leftmost set bit in n (or an upper bound on leftmost position) returns the new position of leftmost set bit in n */ static int getNextLeftmostBit(int n, int m) { int temp = 1 << m; while (n < temp) { temp = temp >> 1; m--; } return m; } // The main recursive function used by countSetBits() // Returns count of set bits present in // all numbers from 1 to n static int countSetBits(int n) { // Get the position of leftmost set // bit in n. This will be used as an // upper bound for next set bit function int m = getLeftmostBit(n); // Use the position return countSetBits(n, m); } static int countSetBits(int n, int m) { // Base Case: if n is 0, then set bit // count is 0 if (n == 0) return 0; /* get position of next leftmost set bit */ m = getNextLeftmostBit(n, m); // If n is of the form 2^x-1, i.e., if n // is like 1, 3, 7, 15, 31, .. etc, // then we are done. // Since positions are considered starting // from 0, 1 is added to m if (n == ((int)1 << (m + 1)) - 1) return (int)(m + 1) * (1 << m); // update n for next recursive call n = n - (1 << m); return (n + 1) + countSetBits(n) + m * (1 << (m - 1)); } // Driver code public static void main(String[] args) { int n = 17; System.out.println("Total set bit count is " + countSetBits(n)); }}// This code is contributed by ajit.. |
Python3
# A O(Logn) complexity program to count# set bits in all numbers from 1 to n"""/* Returns position of leftmost set bit.The rightmost position is consideredas 0 */"""def getLeftmostBit(n): m = 0 while (n > 1): n = n >> 1 m += 1 return m"""/* Given the position of previous leftmostset bit in n (or an upper bound onleftmost position) returns the newposition of leftmost set bit in n */"""def getNextLeftmostBit(n, m): temp = 1 << m while (n < temp): temp = temp >> 1 m -= 1 return m# The main recursive function used by countSetBits()# def _countSetBits(n, m)# Returns count of set bits present in# all numbers from 1 to ndef countSetBits(n): # Get the position of leftmost set # bit in n. This will be used as an # upper bound for next set bit function m = getLeftmostBit(n) # Use the position return _countSetBits(n, m)def _countSetBits(n, m): # Base Case: if n is 0, then set bit # count is 0 if (n == 0): return 0 # /* get position of next leftmost set bit */ m = getNextLeftmostBit(n, m) # If n is of the form 2^x-1, i.e., if n # is like 1, 3, 7, 15, 31, .. etc, # then we are done. # Since positions are considered starting # from 0, 1 is added to m if (n == (1 << (m + 1)) - 1): return ((m + 1) * (1 << m)) # update n for next recursive call n = n - (1 << m) return (n + 1) + countSetBits(n) + m * (1 << (m - 1))# Driver coden = 17print("Total set bit count is", countSetBits(n))# This code is contributed by rathbhupendra |
C#
// C# A O(Logn) complexity program to count// set bits in all numbers from 1 to nusing System;class GFG { /* Returns position of leftmost set bit. The rightmost position is considered as 0 */ static int getLeftmostBit(int n) { int m = 0; while (n > 1) { n = n >> 1; m++; } return m; } /* Given the position of previous leftmost set bit in n (or an upper bound on leftmost position) returns the new position of leftmost set bit in n */ static int getNextLeftmostBit(int n, int m) { int temp = 1 << m; while (n < temp) { temp = temp >> 1; m--; } return m; } // The main recursive function used by countSetBits() // Returns count of set bits present in // all numbers from 1 to n static int countSetBits(int n) { // Get the position of leftmost set // bit in n. This will be used as an // upper bound for next set bit function int m = getLeftmostBit(n); // Use the position return countSetBits(n, m); } static int countSetBits(int n, int m) { // Base Case: if n is 0, // then set bit count is 0 if (n == 0) return 0; /* get position of next leftmost set bit */ m = getNextLeftmostBit(n, m); // If n is of the form 2^x-1, i.e., if n // is like 1, 3, 7, 15, 31, .. etc, // then we are done. // Since positions are considered starting // from 0, 1 is added to m if (n == ((int)1 << (m + 1)) - 1) return (int)(m + 1) * (1 << m); // update n for next recursive call n = n - (1 << m); return (n + 1) + countSetBits(n) + m * (1 << (m - 1)); } // Driver code static public void Main() { int n = 17; Console.Write("Total set bit count is " + countSetBits(n)); }}// This code is contributed by Tushil. |
Javascript
<script>// JavaScript A O(Logn) complexity program to count// set bits in all numbers from 1 to n/* Returns position of leftmost set bit.The rightmost position is consideredas 0 */function getLeftmostBit(n){ let m = 0; while (n > 1) { n = n >> 1; m++; } return m;}/* Given the position of previous leftmostset bit in n (or an upper bound onleftmost position) returns the newposition of leftmost set bit in n */function getNextLeftmostBit(n,m){ let temp = 1 << m; while (n < temp) { temp = temp >> 1; m--; } return m;}// The main recursive function used by countSetBits()// Returns count of set bits present in// all numbers from 1 to nfunction countSetBits(n){ // Get the position of leftmost set // bit in n. This will be used as an // upper bound for next set bit function let m = getLeftmostBit(n); // Use the position return _countSetBits(n, m);}function _countSetBits(n,m){ // Base Case: if n is 0, then set bit // count is 0 if (n == 0) return 0; /* get position of next leftmost set bit */ m = getNextLeftmostBit(n, m); // If n is of the form 2^x-1, i.e., if n // is like 1, 3, 7, 15, 31, .. etc, // then we are done. // Since positions are considered starting // from 0, 1 is added to m if (n == (1 << (m + 1)) - 1) return (m + 1) * (1 << m); // update n for next recursive call n = n - (1 << m); return (n + 1) + countSetBits(n) + m * (1 << (m - 1));}// Driver codelet n = 17;document.write("Total set bit count is " + countSetBits(n)); // This code is contributed by ab2127</script> |
Output
Total set bit count is 35
Time Complexity: O(Logn). From the first look at the implementation, time complexity looks more. But if we take a closer look, statements inside while loop of getNextLeftmostBit() are executed for all 0 bits in n. And the number of times recursion is executed is less than or equal to set bits in n. In other words, if the control goes inside while loop of getNextLeftmostBit(), then it skips those many bits in recursion.
Thanks to agatsu and IC for suggesting this solution.
Here is another solution suggested by Piyush Kapoor.
A simple solution , using the fact that for the ith least significant bit, answer will be
(N/2^i)*2^(i-1)+ X
where
X = N%(2^i)-(2^(i-1)-1)
if
N%(2^i)>(2^(i-1)-1)
C++
int getSetBitsFromOneToN(int N){ int two = 2, ans = 0; int n = N; while (n) { ans += (N / two) * (two >> 1); if ((N & (two - 1)) > (two >> 1) - 1) ans += (N & (two - 1)) - (two >> 1) + 1; two <<= 1; n >>= 1; } return ans;} |
Java
static int getSetBitsFromOneToN(int N){ int two = 2, ans = 0; int n = N; while (n != 0) { ans += (N / two) * (two >> 1); if ((N & (two - 1)) > (two >> 1) - 1) ans += (N & (two - 1)) - (two >> 1) + 1; two <<= 1; n >>= 1; } return ans;}// This code is contributed by divyeshrabadiya07. |
Python3
def getSetBitsFromOneToN(N): two = 2 ans = 0 n = N while(n != 0) { ans += int(N / two) * (two >> 1) if((N & (two - 1)) > (two >> 1) - 1): ans += (N & (two - 1)) - (two >> 1) + 1 two <<= 1 n >>= 1 } return ans# This code is contributed by avanitrachhadiya2155 |
C#
static int getSetBitsFromOneToN(int N){ int two = 2, ans = 0; int n = N; while (n != 0) { ans += (N / two) * (two >> 1); if ((N & (two - 1)) > (two >> 1) - 1) ans += (N & (two - 1)) - (two >> 1) + 1; two <<= 1; n >>= 1; } return ans;}// This code is contributed by divyesh072019. |
Javascript
<script> function getSetBitsFromOneToN(N){ var two = 2 var ans = 0 var n = N while(n != 0) { ans += Math.floor(N / two) * (two >> 1) if ((N & (two - 1)) > (two >> 1) - 1) ans += (N&(two - 1)) - (two >> 1) + 1 two <<= 1; n >>= 1; } return ans}// This code is contributed by AnkThon</script> |
Method 4 (Recursive)
Approach:
For each number ‘n’, there will be a number a, where a<=n and a is perfect power of two, like 1,2,4,8…..
Let n = 11, now we can see that
Numbers till n, are:
0 -> 0000000
1 -> 0000001
2 -> 0000010
3 -> 0000011
4 -> 0000100
5 -> 0000101
6 -> 0000110
7 -> 0000111
8 -> 0001000
9 -> 0001001
10 -> 0001010
11 -> 0001011
Now we can see that, from 0 to pow(2,1)-1 = 1, we can pair elements top-most with bottom-most,
and count of set bit in a pair is 1
Similarly for pow(2,2)-1 = 4, pairs are:
00 and 11
01 and 10
here count of set bit in a pair is 2, so in both pairs is 4
Similarly we can see for 7, 15, ans soon.....
so we can generalise that,
count(x) = (x*pow(2,(x-1))),
here x is position of set bit of the largest power of 2 till n
for n = 8, x = 3
for n = 4, x = 2
for n = 5, x = 2
so now for n = 11,
we have added set bits count from 0 to 7 using count(x) = (x*pow(2,(x-1)))
for rest numbers 8 to 11, all will have a set bit at 3rd index, so we can add
count of rest numbers to our ans,
which can be calculated using 11 - 8 + 1 = (n-pow(2,x) + 1)
Now if notice that, after removing front bits from rest numbers, we get again number from 0 to some m
so we can recursively call our same function for next set of numbers,
by calling countSetBits(n - pow(2,x))
8 -> 1000 -> 000 -> 0
9 -> 1001 -> 001 -> 1
10 -> 1010 -> 010 -> 2
11 -> 1011 -> 011 -> 3
Code:
C++
#include <bits/stdc++.h>using namespace std;int findLargestPower(int n){ int x = 0; while ((1 << x) <= n) x++; return x - 1;}int countSetBits(int n){ if (n <= 1) return n; int x = findLargestPower(n); return (x * pow(2, (x - 1))) + (n - pow(2, x) + 1) + countSetBits(n - pow(2, x));}int main(){ int N = 17; cout << countSetBits(N) << endl; return 0;} |
Java
import java.io.*;class GFG { static int findLargestPower(int n) { int x = 0; while ((1 << x) <= n) x++; return x - 1; } static int countSetBits(int n) { if (n <= 1) return n; int x = findLargestPower(n); return (x * (int)Math.pow(2, (x - 1))) + (n - (int)Math.pow(2, x) + 1) + countSetBits(n - (int)Math.pow(2, x)); } public static void main(String[] args) { int N = 17; System.out.print(countSetBits(N)); }}// This code is contributed by shivanisinghss2110 |
Python3
def findLargestPower(n): x = 0 while ((1 << x) <= n): x += 1 return x - 1def countSetBits(n): if (n <= 1): return n x = findLargestPower(n) return (x * pow(2, (x - 1))) + (n - pow(2, x) + 1) + countSetBits(n - pow(2, x))N = 17print(countSetBits(N))# This code is contributed by shivanisinghss2110 |
C#
using System;class GFG { static int findLargestPower(int n) { int x = 0; while ((1 << x) <= n) x++; return x - 1; } static int countSetBits(int n) { if (n <= 1) return n; int x = findLargestPower(n); return (x * (int)Math.Pow(2, (x - 1))) + (n - (int)Math.Pow(2, x) + 1) + countSetBits(n - (int)Math.Pow(2, x)); } public static void Main(string[] args) { int N = 17; Console.Write(countSetBits(N)); }}// This code is contributed by shivanisinghss2110 |
Javascript
<script>function findLargestPower(n){ var x = 0; while ((1 << x) <= n) x++; return x - 1;}function countSetBits(n){ if (n <= 1) return n; var x = findLargestPower(n); return (x * Math.pow(2, (x - 1))) + (n - Math.pow(2, x) + 1) + countSetBits(n - Math.pow(2, x));}var N = 17;document.write(countSetBits(N));// this code is contributed by shivanisinghss2110</script> |
Output
35
Time Complexity: O(LogN)
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
Method 5(Iterative)
Example:
0 -> 0000000 8 -> 001000 16 -> 010000 24 -> 011000
1 -> 0000001 9 -> 001001 17 -> 010001 25 -> 011001
2 -> 0000010 10 -> 001010 18 -> 010010 26 -> 011010
3 -> 0000011 11 -> 001011 19 -> 010010 27 -> 011011
4 -> 0000100 12 -> 001100 20 -> 010100 28 -> 011100
5 -> 0000101 13 -> 001101 21 -> 010101 29 -> 011101
6 -> 0000110 14 -> 001110 22 -> 010110 30 -> 011110
7 -> 0000111 15 -> 001111 23 -> 010111 31 -> 011111
Input: N = 4
Output: 5
Input: N = 17
Output: 35
Approach : Pattern recognition
Let ‘N’ be any arbitrary number and consider indexing from right to left(rightmost being 1); then nearestPow = pow(2,i).
Now, when you write all numbers from 1 to N, you will observe the pattern mentioned below:
For every index i, there are exactly nearestPow/2 continuous elements that are unset followed by nearestPow/2 elements that are set.
Throughout the solution, i am going to use this concept.
You can clearly observe above concept in the above table.
The general formula that i came up with:
a. addRemaining = mod – (nearestPos/2) + 1 if mod >= nearestPow/2;
b. totalSetBitCount = totalRep*(nearestPow/2) + addRemaining
where totalRep -> total number of times the pattern repeats at index i
addRemaining -> total number of set bits left to be added after the pattern is exhausted
Eg: let N = 17
leftMostSetIndex = 5 (Left most set bit index, considering 1 based indexing)
i = 1 => nearestPos = pow(2,1) = 2; totalRep = (17+1)/2 = 9 (add 1 only for i=1)
mod = 17%2 = 1
addRemaining = 0 (only for base case)
totalSetBitCount = totalRep*(nearestPos/2) + addRemaining = 9*(2/2) + 0 = 9*1 + 0 = 9
i = 2 => nearestPos = pow(2, 2)=4; totalRep = 17/4 = 4
mod = 17%4 = 1
mod(1) < (4/2) => 1 < 2 => addRemaining = 0
totalSetBitCount = 9 + 4*(2) + 0 = 9 + 8 = 17
i = 3 => nearestPow = pow(2,3) = 8; totalRep = 17/8 = 2
mod = 17%8 = 1
mod < 4 => addRemaining = 0
totalSetBitCount = 17 + 2*(4) + 0 = 17 + 8 + 0 = 25
i = 4 => nearestPow = pow(2, 4) = 16; totalRep = 17/16 = 1
mod = 17%16 = 1
mod < 8 => addRemaining = 0
totalSetBitCount = 25 + 1*(8) + 0 = 25 + 8 + 0 = 33
We cannot simply operate on the next power(32) as 32>17. Also, as the first half bits will be 0s only, we need to find the distance of the given number(17) from the last power to directly get the number of 1s to be added
i = 5 => nearestPow = (2, 5) = 32 (base case 2)
lastPow = pow(2, 4) = 16
mod = 17%16 = 1
totalSetBit = 33 + (mod+1) = 33 + 1 + 1 = 35
Therefore, total num of set bits from 1 to 17 is 35
Try iterating with N = 30, for better understanding of the solution.
Solution
C++
#include <bits/stdc++.h>#include <iostream>using namespace std;int GetLeftMostSetBit(int n){ int pos = 0; while (n > 0) { pos++; n >>= 1; } return pos;}int TotalSetBitsFrom1ToN(int n){ int leftMostSetBitInd = GetLeftMostSetBit(n); int totalRep, mod; int nearestPow; int totalSetBitCount = 0; int addRemaining = 0; int curr = 0; // denotes the number of set bits at index i // cout<<"leftMostSetBitInd: "<<leftMostSetBitInd<<endl; for (int i = 1; i <= leftMostSetBitInd; ++i) { nearestPow = pow(2, i); if (nearestPow > n) { int lastPow = pow(2, i - 1); mod = n % lastPow; totalSetBitCount += mod + 1; } else { if (i == 1 && n % 2 == 1) { totalRep = (n + 1) / nearestPow; mod = nearestPow % 2; addRemaining = 0; } else { totalRep = n / nearestPow; mod = n % nearestPow; if (mod >= (nearestPow / 2)) { addRemaining = mod - (nearestPow / 2) + 1; } else { addRemaining = 0; } } curr = totalRep * (nearestPow / 2) + addRemaining; totalSetBitCount += curr; } // debug output at each iteration // cout<<i<<" "<<nearestPow<<" "<<totalRep<<" // "<<mod<<" "<<totalSetBitCount<<" "<<curr<<endl; } return totalSetBitCount;}int main(){ std::cout << TotalSetBitsFrom1ToN(4) << endl; std::cout << TotalSetBitsFrom1ToN(17) << endl; std::cout << TotalSetBitsFrom1ToN(30) << endl; return 0;}// This code is contributed by rjkrsngh |
Java
import java.io.*;import java.util.*;class GFG { static int GetLeftMostSetBit(int n) { int pos = 0; while (n > 0) { pos++; n >>= 1; } return pos; } static int TotalSetBitsFrom1ToN(int n) { int leftMostSetBitInd = GetLeftMostSetBit(n); int totalRep, mod; int nearestPow; int totalSetBitCount = 0; int addRemaining = 0; int curr = 0; // denotes the number of set bits at // index i for (int i = 1; i <= leftMostSetBitInd; ++i) { nearestPow = (int)Math.pow(2, i); if (nearestPow > n) { int lastPow = (int)Math.pow(2, i - 1); mod = n % lastPow; totalSetBitCount += mod + 1; } else { if (i == 1 && n % 2 == 1) { totalRep = (n + 1) / nearestPow; mod = nearestPow % 2; addRemaining = 0; } else { totalRep = n / nearestPow; mod = n % nearestPow; if (mod >= (nearestPow / 2)) { addRemaining = mod - (nearestPow / 2) + 1; } else { addRemaining = 0; } } curr = totalRep * (nearestPow / 2) + addRemaining; totalSetBitCount += curr; } } return totalSetBitCount; } public static void main(String[] args) { System.out.println(TotalSetBitsFrom1ToN(4)); System.out.println(TotalSetBitsFrom1ToN(17)); System.out.println(TotalSetBitsFrom1ToN(30)); }}// This code is contributed by vikaschhonkar1// By: Vikas Chhonkar |
Python3
def get_left_most_set_bit(n): left_most_set_bit_indx = 0 while n > 0: left_most_set_bit_indx += 1 n >>= 1 return left_most_set_bit_indxdef total_set_bits_from_1_to_n(n): left_most_set_bit_indx = get_left_most_set_bit(n) total_rep = 0 mod = 0 nearest_pow = 0 total_set_bit_count = 0 add_remaining = 0 curr = 0 # denotes the number of set bits at index i for i in range(1, left_most_set_bit_indx + 1): nearest_pow = pow(2, i) if nearest_pow > n: last_pow = pow(2, i-1) mod = n % last_pow total_set_bit_count += mod + 1 else: if i == 1 and n % 2 == 1: total_rep = (n + 1) / nearest_pow mod = nearest_pow % 2 add_remaining = 0 else: total_rep = int(n / nearest_pow) mod = n % nearest_pow add_remaining = int( mod - (nearest_pow / 2) + 1) if mod >= nearest_pow / 2 else 0 curr = int(total_rep * (nearest_pow / 2) + add_remaining) total_set_bit_count += curr return total_set_bit_count# Driver codeif __name__ == "__main__": print(total_set_bits_from_1_to_n(4)) print(total_set_bits_from_1_to_n(17)) print(total_set_bits_from_1_to_n(30)) # This code is contributed by tssovi. |
C#
// C# program for the above approachusing System;public class GFG { static int GetLeftMostSetBit(int n) { int pos = 0; while (n > 0) { pos++; n >>= 1; } return pos; } static int TotalSetBitsFrom1ToN(int n) { int leftMostSetBitInd = GetLeftMostSetBit(n); int totalRep, mod; int nearestPow; int totalSetBitCount = 0; int addRemaining = 0; int curr = 0; // denotes the number of set bits at // index i for (int i = 1; i <= leftMostSetBitInd; ++i) { nearestPow = (int)Math.Pow(2, i); if (nearestPow > n) { int lastPow = (int)Math.Pow(2, i - 1); mod = n % lastPow; totalSetBitCount += mod + 1; } else { if (i == 1 && n % 2 == 1) { totalRep = (n + 1) / nearestPow; mod = nearestPow % 2; addRemaining = 0; } else { totalRep = n / nearestPow; mod = n % nearestPow; if (mod >= (nearestPow / 2)) { addRemaining = mod - (nearestPow / 2) + 1; } else { addRemaining = 0; } } curr = totalRep * (nearestPow / 2) + addRemaining; totalSetBitCount += curr; } } return totalSetBitCount; } // Driver Code public static void Main(String[] args) { Console.WriteLine(TotalSetBitsFrom1ToN(4)); Console.WriteLine(TotalSetBitsFrom1ToN(17)); Console.WriteLine(TotalSetBitsFrom1ToN(30)); }}// This code is contributed by code_hunt. |
Javascript
<script> function GetLeftMostSetBit(n) { var pos = 0; while (n > 0) { pos++; n >>= 1; } return pos; } function TotalSetBitsFrom1ToN(n) { var leftMostSetBitInd = GetLeftMostSetBit(n); var totalRep, mod; var nearestPow; var totalSetBitCount = 0; var addRemaining = 0; var curr = 0; // denotes the number of set bits at index i for (var i = 1; i <= leftMostSetBitInd; ++i) { nearestPow = parseInt( Math.pow(2, i)); if (nearestPow > n) { var lastPow = parseInt( Math.pow(2, i - 1)); mod = n % lastPow; totalSetBitCount += mod + 1; } else { if (i == 1 && n % 2 == 1) { totalRep = parseInt((n + 1) / nearestPow); mod = nearestPow % 2; addRemaining = 0; } else { totalRep =parseInt( n / nearestPow); mod = n % nearestPow; if (mod >= parseInt(nearestPow / 2)) { addRemaining = mod - parseInt(nearestPow / 2) + 1; } else { addRemaining = 0; } } curr = totalRep * (parseInt(nearestPow / 2)) + addRemaining; totalSetBitCount += curr; } } return totalSetBitCount; } document.write(TotalSetBitsFrom1ToN(4)); document.write("<br/>"+TotalSetBitsFrom1ToN(17)); document.write("<br/>"+TotalSetBitsFrom1ToN(30));// This code is contributed by Rajput-Ji </script> |
Output
5 35 75
Time Complexity: O(log(n))
Method 6 ( Using Inbuilt Functions )
__builtin_popcount(): Counts a number of set bits in a number.
C++
#include <bits/stdc++.h>#include <iostream>int TotalSetBitsFrom1ToN(int n){ int totalSetBitCount = 0; for(int number = 1 ; number <= n ;number++) { totalSetBitCount+=__builtin_popcount(number); } return totalSetBitCount;}int main(){ std::cout << TotalSetBitsFrom1ToN(4) << std::endl; std::cout << TotalSetBitsFrom1ToN(17) << std::endl; std::cout << TotalSetBitsFrom1ToN(30) << std::endl; return 0;}// This code is contributed by yash_cse_21 |
Java
public class GFG { public static int totalSetBitsFrom1ToN(int n) { int totalSetBitCount = 0; for (int number = 1; number <= n; number++) { totalSetBitCount += Integer.bitCount(number); } return totalSetBitCount; } public static void main(String[] args) { System.out.println(totalSetBitsFrom1ToN(4)); System.out.println(totalSetBitsFrom1ToN(17)); System.out.println(totalSetBitsFrom1ToN(30)); }} |
Python3
def total_set_bits_from_1_to_n(n): total_set_bit_count = 0 # Loop through numbers from 1 to n (inclusive) for number in range(1, n+1): # Convert the number to its binary representation and count the set bits (ones) total_set_bit_count += bin(number).count('1') # Return the total count of set bits from 1 to n return total_set_bit_countdef main(): # Test cases to demonstrate the function print("Total set bits from 1 to 4:", total_set_bits_from_1_to_n(4)) print("Total set bits from 1 to 17:", total_set_bits_from_1_to_n(17)) print("Total set bits from 1 to 30:", total_set_bits_from_1_to_n(30))if __name__ == "__main__": main()# This code is contributed by akshitaguprzj3 |
C#
using System;class GFG{ // Function to count the total number of set bits in numbers from 1 to n static int TotalSetBitsFrom1ToN(int n) { int totalSetBitCount = 0; // Iterate through numbers from 1 to n for (int number = 1; number <= n; number++) { // Count the set bits in the current number and add to the total count totalSetBitCount += BitCount(number); } return totalSetBitCount; } // Function to count the number of set bits in a given number static int BitCount(int number) { int count = 0; // Perform bitwise AND operations to count set bits while (number > 0) { count += number & 1; // Increment count if the least significant bit is set number >>= 1; // Right shift number to check the next bit } return count; } static void Main(string[] args) { // Calculate and display the total set bit count for different input values Console.WriteLine(TotalSetBitsFrom1ToN(4)); Console.WriteLine(TotalSetBitsFrom1ToN(17)); Console.WriteLine(TotalSetBitsFrom1ToN(30)); }} |
Javascript
function totalSetBitsFrom1ToN(n) { let totalSetBitCount = 0; for (let number = 1; number <= n; number++) { totalSetBitCount += number.toString(2).split('1').length - 1; } return totalSetBitCount;}console.log(totalSetBitsFrom1ToN(4));console.log(totalSetBitsFrom1ToN(17));console.log(totalSetBitsFrom1ToN(30)); |
Output
5 35 75
Time Complexity: O(n*k) ,where n is total numbers and k is total number of bits in the number.
Space Complexity: O(1)
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