Given an array arr (1-based indexing) of length N and an integer X, the task is to find and print all index ranges having a set bit sum equal to X in the array.
Examples:
Input: A[] = {1 4 3 5 7}, X = 4
Output: (1, 3), (3, 4)
Explanation: In the above array subarray having set bit sum equal to X (= 4).
Starting from index 1 to 3. {1 4 3} = (001) + (100) + (011) = 4 and
other one is from 3 to 4 {3, 5} = (011) + (101) = 4.Input: arr[] = {5, 3, 0, 4, 10}, X = 7
Output: (1 5)
Explanation: In the above array subarrays having set bit sum equal to X(= 7) start from 1 to 5 only.
Approach: The problem is solved using two pointer approach.
- Write a function countSetBit to count the number of set bits.
- Initialize a counter c=0, to store the individual count for every number in the array.
- Iterate over the array and check for every set bit and increase the counter.
- replace every number with the count of a number of set bits
- Write a function to print a range of subarrays PrintIndex
Run a loop using two pointers i and j and check for the sum as follow:- If the current index sum is less than X then, add the value at arr[j] in currsum
- else if the sum is equal to X push back the start and end index of the array and increment the counter i.
- else decrement the counter, subtract the value at arr[i] from currsum.
- Repeat the same for all elements.
Below is the implementation of the above method :
C++
// C++ program to Find all range// Having set bit sum X in array#include <bits/stdc++.h>using namespace std;// Function to replace elements// With their set bit countvoid countSetBit(vector<int>& arr, int n){ int c = 0, i; for (i = 0; i < n; i++) { int x = arr[i]; while (x) { int l = x % 10; if (x & 1) c++; x /= 2; } // Replace array element // to set bit count arr[i] = c; c = 0; }}// Function to find range of subarrays// having set bit sum equal to X.void PrintIndex(vector<int> arr, int N, int X, vector<int>& v){ int i = 0, j = 0, currSum = arr[0]; while (j < N && i < N) { if (currSum == X) { // push back index i start // point ans end point j // when sum == X v.push_back(i + 1); v.push_back(j + 1); j++; currSum += arr[j]; } // when current sum is // less than X increment j // and add arr[j] else if (currSum < X) { j++; currSum += arr[j]; } // when current sum is // greater than X increment j // and subtract arr[i] else { currSum -= arr[i]; i++; } }}// Driver codeint main(){ vector<int> v = { 1, 4, 3, 5, 7 }; int X = 4; int N = v.size(); // replace all the array element into // their set bit count value countSetBit(v, N); vector<int> ans; PrintIndex(v, N, X, ans); for (int i = 0; i < ans.size() - 1; i += 2) cout << "(" << ans[i] << " " << ans[i + 1] << ")" << " "; return 0;} |
Java
// JAVA code to implement the above approachimport java.util.*;class GFG { // Function to replace elements // With their set bit count static void countSetBit(int[] arr, int n) { int c = 0, i; for (i = 0; i < n; i++) { int x = arr[i]; while (x > 0) { int l = x % 10; if ((x & 1) == 1) c++; x /= 2; } // Replace array element // to set bit count arr[i] = c; c = 0; } } // Function to find range of subarrays // having set bit sum equal to X. static void PrintIndex(int[] arr, int N, int X, ArrayList<Integer> v) { int i = 0, j = 0, currSum = arr[0]; while (j < N && i < N) { if (currSum == X) { // push back index i start // point ans end point j // when sum == X v.add(i + 1); v.add(j + 1); j++; if (j < N) currSum += arr[j]; } // when current sum is // less than X increment j // and add arr[j] else if (currSum < X) { j++; if (j < N) currSum += arr[j]; } // when current sum is // greater than X increment j // and subtract arr[i] else { currSum -= arr[i]; i++; } } } // Driver Code public static void main(String[] args) { int[] v = { 1, 4, 3, 5, 7 }; int X = 4; int N = v.length; // replace all the array element into // their set bit count value countSetBit(v, N); ArrayList<Integer> ans = new ArrayList<Integer>(); PrintIndex(v, N, X, ans); for (int i = 0; i < ans.size() - 1; i += 2) System.out.print("(" + ans.get(i) + " " + ans.get(i + 1) + ")" + " "); }}// This code is contributed by sanjoy_62. |
Python3
# Python program to Find all range# Having set bit sum X in array# Function to replace elements# With their set bit countdef countSetBit(arr, n): c = 0 for i in range(n): x = arr[i] while (x): l = x % 10 if (x & 1): c += 1 x = x // 2 # Replace array element # to set bit count arr[i] = c c = 0# Function to find range of subarrays# having set bit sum equal to X.def PrintIndex(arr, N, X, v): i,j,currSum = 0,0,arr[0] while (j < N and i < N): if (currSum == X): # append back index i start # point ans end point j # when sum == X v.append(i + 1) v.append(j + 1) j += 1 if(j<N): currSum += arr[j] # when current sum is # less than X increment j # and add arr[j] elif (currSum < X): j += 1 if(j<N): currSum += arr[j] # when current sum is # greater than X increment j # and subtract arr[i] else: currSum -= arr[i] i += 1# Driver codev = [1, 4, 3, 5, 7]X = 4N = len(v)# replace all the array element into# their set bit count valuecountSetBit(v, N)ans = []PrintIndex(v, N, X, ans)for i in range(0,len(ans) - 1,2): print(f"({ans[i]} {ans[i + 1]})",end=" ")# This code is contributed by shinjanpatra |
C#
// C# program to Find all range// Having set bit sum X in arrayusing System;using System.Collections;class GFG { // Function to replace elements // With their set bit count static void countSetBit(int[] arr, int n) { int c = 0, i; for (i = 0; i < n; i++) { int x = arr[i]; while (x > 0) { int l = x % 10; if ((x & 1) == 1) c++; x /= 2; } // Replace array element // to set bit count arr[i] = c; c = 0; } } // Function to find range of subarrays // having set bit sum equal to X. static void PrintIndex(int[] arr, int N, int X, ArrayList v) { int i = 0, j = 0, currSum = arr[0]; while (j < N && i < N) { if (currSum == X) { // push back index i start // point ans end point j // when sum == X v.Add(i + 1); v.Add(j + 1); j++; if (j < N) currSum += arr[j]; } // when current sum is // less than X increment j // and add arr[j] else if (currSum < X) { j++; if (j < N) currSum += arr[j]; } // when current sum is // greater than X increment j // and subtract arr[i] else { currSum -= arr[i]; i++; } } } // Driver code public static void Main() { int[] v = { 1, 4, 3, 5, 7 }; int X = 4; int N = v.Length; // replace all the array element into // their set bit count value countSetBit(v, N); ArrayList ans = new ArrayList(); PrintIndex(v, N, X, ans); for (int i = 0; i < ans.Count - 1; i += 2) Console.Write("(" + ans[i] + " " + ans[i + 1] + ")" + " "); }}// This code is contributed by Samim Hossain Mondal. |
Javascript
<script> // JavaScript program to Find all range // Having set bit sum X in array // Function to replace elements // With their set bit count const countSetBit = (arr, n) => { let c = 0, i; for (i = 0; i < n; i++) { let x = arr[i]; while (x) { let l = x % 10; if (x & 1) c++; x = parseInt(x / 2); } // Replace array element // to set bit count arr[i] = c; c = 0; } } // Function to find range of subarrays // having set bit sum equal to X. const PrintIndex = (arr, N, X, v) => { let i = 0, j = 0, currSum = arr[0]; while (j < N && i < N) { if (currSum == X) { // push back index i start // point ans end point j // when sum == X v.push(i + 1); v.push(j + 1); j++; currSum += arr[j]; } // when current sum is // less than X increment j // and add arr[j] else if (currSum < X) { j++; currSum += arr[j]; } // when current sum is // greater than X increment j // and subtract arr[i] else { currSum -= arr[i]; i++; } } } // Driver code let v = [1, 4, 3, 5, 7]; let X = 4; let N = v.length; // replace all the array element into // their set bit count value countSetBit(v, N); let ans = []; PrintIndex(v, N, X, ans); for (let i = 0; i < ans.length - 1; i += 2) document.write(`(${ans[i]} ${ans[i + 1]}) `);// This code is contributed by rakeshsahni</script> |
Output
(1 3) (3 4)
Time Complexity: O(N * d) where d is the count of bits in an array element
Auxiliary Space: O(N)
Another Approach:
- The code defines a function called countSetBit that takes an integer x and returns the number of set bits in its binary representation.
- The code also defines another function called printSubarraysWithSetBitSumX that takes a vector of integers arr and an integer X. This function prints all the subarrays of arr whose sum of set bits is equal to X.
- Inside the printSubarraysWithSetBitSumX function, the code initializes some variables: n is the size of the input vector arr, i and j are two pointers initially set to 0, and currSum is the current sum of set bits.
- The code enters a while loop with a condition of j < n. This loop iterates through all the elements of the input vector arr.
- Inside the while loop, the code adds the count of set bits of the current element to the currSum variable and increments j by 1.
- The code then enters another while loop with a condition of currSum > X. This loop removes the set bit count of the element pointed by i from the currSum variable and increments i by 1 until the currSum becomes less than or equal to X.
- If the currSum is equal to X after the above while loop, the code prints the indices of the subarray whose sum of set bits is equal to X.
- The while loop in step 4 continues until j reaches the end of the input vector arr.
- Finally, the main function creates a vector arr containing integers {1, 4, 3, 5, 7} and sets X to 4. It then calls the printSubarraysWithSetBitSumX function with these arguments, which prints “(1, 3) (3, 4)” to the console.
Below is the implementation of the above approach:
C++
#include <iostream>#include <vector>using namespace std;int countSetBit(int x) { int count = 0; while (x > 0) { count += x & 1; x >>= 1; } return count;}void printSubarraysWithSetBitSumX(vector<int>& arr, int X) { int n = arr.size(); int i = 0, j = 0, currSum = 0; while (j < n) { currSum += countSetBit(arr[j]); j++; while (currSum > X) { currSum -= countSetBit(arr[i]); i++; } if (currSum == X) { cout << "(" << i + 1 << ", " << j << ") "; } }}int main() { vector<int> arr = {1, 4, 3, 5, 7}; int X = 4; printSubarraysWithSetBitSumX(arr, X); // prints (1, 3) (3, 4) return 0;} |
Java
import java.util.*;public class SubarraysWithSetBitSum { public static int countSetBit(int x) { int count = 0; while (x > 0) { count += x & 1; x >>= 1; } return count; } public static void printSubarraysWithSetBitSumX(ArrayList<Integer> arr, int X) { int n = arr.size(); int i = 0, j = 0, currSum = 0; while (j < n) { currSum += countSetBit(arr.get(j)); j++; while (currSum > X) { currSum -= countSetBit(arr.get(i)); i++; } if (currSum == X) { System.out.print("(" + (i + 1) + ", " + j + ") "); } } } public static void main(String[] args) { ArrayList<Integer> arr = new ArrayList<Integer>(Arrays.asList(1, 4, 3, 5, 7)); int X = 4; printSubarraysWithSetBitSumX(arr, X); // prints (1, 3) (3, 4) }} |
Python3
def count_set_bits(x): # Function to count the number of set bits (1s) in a binary representation of 'x' count = 0 while x > 0: count += x & 1 # Add the least significant bit of 'x' to the count x >>= 1 # Right shift 'x' to check the next bit return countdef print_subarrays_with_set_bit_sum_x(arr, X): n = len(arr) # Get the length of the input array 'arr' i, j, curr_sum = 0, 0, 0 # Initialize variables for subarray tracking while j < n: # Iterate through the array using a sliding window (j moves forward) curr_sum += count_set_bits(arr[j]) # Add the count of set bits in 'arr[j]' to 'curr_sum' j += 1 while curr_sum > X: # If 'curr_sum' exceeds the target 'X', move the window's left end (i) forward curr_sum -= count_set_bits(arr[i]) # Subtract the count of set bits in 'arr[i]' from 'curr_sum' i += 1 if curr_sum == X: # If 'curr_sum' matches the target 'X', print the subarray print(f"({i + 1}, {j}) ", end='')# Main functionif __name__ == "__main__": arr = [1, 4, 3, 5, 7] # Input array X = 4 # Target sum of set bits print_subarrays_with_set_bit_sum_x(arr, X) # Call the function to find and print subarrays with the target sum |
C#
using System;using System.Collections.Generic;public class SubarraysWithSetBitSum { // Function to count the number of set bits in a number public static int CountSetBit(int x) { int count = 0; while (x > 0) { count += x & 1; // Increment count if the last bit is 1 x >>= 1; // Right shift the number to check the next bit } return count; } // Function to find and print subarrays with a given set bit sum public static void PrintSubarraysWithSetBitSumX(List<int> arr, int X) { int n = arr.Count; int i = 0, j = 0, currSum = 0; while (j < n) { currSum += CountSetBit(arr[j]); // Calculate set bit sum for the current element j++; // Slide the window to the right while current set bit sum is greater than X while (currSum > X) { currSum -= CountSetBit(arr[i]); // Remove set bit count of the left element i++; } // If the current set bit sum matches X, print the subarray indices if (currSum == X) { Console.Write("(" + (i + 1) + ", " + j + ") "); } } } public static void Main(string[] args) { List<int> arr = new List<int> { 1, 4, 3, 5, 7 }; int X = 4; PrintSubarraysWithSetBitSumX(arr, X); // prints (1, 3) (3, 4) }} |
Javascript
function countSetBit(x) { let count = 0; while (x > 0) { count += x & 1; x >>= 1; } return count;}function printSubarraysWithSetBitSumX(arr, X) { const n = arr.length; let i = 0, j = 0, currSum = 0; while (j < n) { currSum += countSetBit(arr[j]); j++; while (currSum > X) { currSum -= countSetBit(arr[i]); i++; } if (currSum === X) { console.log(`(${i + 1}, ${j})`); } }}const arr = [1, 4, 3, 5, 7];const X = 4;printSubarraysWithSetBitSumX(arr, X); // prints (1, 3) (3, 4) |
Output
(1, 3) (3, 4)
Time complexity: O(n*logx)
Auxiliary Space: O(n)
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