We are given two numbers n and m, and two-bit positions, i and j. Insert bits of m into n starting from j to i. We can assume that the bits j through i have enough space to fit all of m. That is, if m = 10011, you can assume that there are at least 5 bits between j and i. You would not, for example, have j = 3 and i = 2, because m could not fully fit between bit 3 and bit 2.
Examples :
Input : n = 1024
m = 19
i = 2
j = 6;
Output : n = 1100
Binary representations of input numbers
m in binary is (10011)2
n in binary is (10000000000)2
Binary representations of output number
(10000000000)2
Input : n = 5
m = 3
i = 1
j = 2
Output : 7
Algorithm :
1. Clear the bits j through i in n 2. Shift m so that it lines up with bits j through i 3. Return Bitwise AND of m and n.
The trickiest part is Step 1. How do we clear the bits in n? We can do this with a mask. This mask will have all 1s, except for 0s in the bits j through i. We create this mask by creating the left half of the mask first, and then the right half.
Following is the implementation of the above approach.
C++
// C++ program for implementation of updateBits()#include <bits/stdc++.h>using namespace std;// Function to updateBits M insert to N.int updateBits(int n, int m, int i, int j){ /* Create a mask to clear bits i through j in n. EXAMPLE: i = 2, j = 4. Result should be 11100011. For simplicity, we'll use just 8 bits for the example. */ int allOnes = ~0; // will equal sequence of all ls // ls before position j, then 0s. left = 11100000 int left= allOnes << (j + 1); // l's after position i. right = 00000011 int right = ((1 << i) - 1); // All ls, except for 0s between i and j. mask 11100011 int mask = left | right; /* Clear bits j through i then put min there */ int n_cleared = n & mask; // Clear bits j through i. int m_shifted = m << i; // Move m into correct position. return (n_cleared | m_shifted); // OR them, and we're done!}// Driver Codeint main(){ int n = 1024; // in Binary N= 10000000000 int m = 19; // in Binary M= 10011 int i = 2, j = 6; cout << updateBits(n,m,i,j); return 0;} |
Java
// Java program for implementation of updateBits()class UpdateBits { // Function to updateBits M insert to N. static int updateBits(int n, int m, int i, int j) { /* Create a mask to clear bits i through j in n. EXAMPLE: i = 2, j = 4. Result should be 11100011. For simplicity, we'll use just 8 bits for the example. */ int allOnes = ~0; // will equal sequence of all ls // ls before position j, then 0s. left = 11100000 int left= allOnes << (j + 1); // l's after position i. right = 00000011 int right = ((1 << i) - 1); // All ls, except for 0s between i and j. mask 11100011 int mask = left | right; /* Clear bits j through i then put min there */ // Clear bits j through i. int n_cleared = n & mask; // Move m into correct position. int m_shifted = m << i; // OR them, and we're done! return (n_cleared | m_shifted); } public static void main (String[] args) { // in Binary N= 10000000000 int n = 1024; // in Binary M= 10011 int m = 19; int i = 2, j = 6; System.out.println(updateBits(n,m,i,j)); }} |
Python3
# Python3 program for implementation# of updateBits()# Function to updateBits M insert to N.def updateBits(n, m, i, j): # Create a mask to clear bits i through # j in n. EXAMPLE: i = 2, j = 4. Result # should be 11100011. For simplicity, # we'll use just 8 bits for the example. # will equal sequence of all ls allOnes = ~0 # ls before position j, # then 0s. left = 11100000 left = allOnes << (j + 1) # l's after position i. right = 00000011 right = ((1 << i) - 1) # All ls, except for 0s between # i and j. mask 11100011 mask = left | right # Clear bits j through i then put min there n_cleared = n & mask # Move m into correct position. m_shifted = m << i return (n_cleared | m_shifted)# Driver Coden = 1024 # in Binary N = 10000000000m = 19 # in Binary M = 10011i = 2; j = 6print(updateBits(n, m, i, j))# This code is contributed by Anant Agarwal. |
C#
// C# program for implementation of // updateBits()using System;class GFG { // Function to updateBits M // insert to N. static int updateBits(int n, int m, int i, int j) { /* Create a mask to clear bits i through j in n. EXAMPLE: i = 2, j = 4. Result should be 11100011. For simplicity, we'll use just 8 bits for the example. */ // will equal sequence of all ls int allOnes = ~0; // ls before position j, then 0s. // left = 11100000 int left= allOnes << (j + 1); // l's after position i. // right = 00000011 int right = ((1 << i) - 1); // All ls, except for 0s between i // and j. mask 11100011 int mask = left | right; /* Clear bits j through i then put min there */ // Clear bits j through i. int n_cleared = n & mask; // Move m into correct position. int m_shifted = m << i; // OR them, and we're done! return (n_cleared | m_shifted); } public static void Main() { // in Binary N= 10000000000 int n = 1024; // in Binary M= 10011 int m = 19; int i = 2, j = 6; Console.WriteLine(updateBits(n, m, i, j)); }}//This code is contributed by Anant Agarwal. |
PHP
<?php// PHP program for implementation // of updateBits()// Function to updateBits // M insert to N.function updateBits($n, $m, $i, $j){ // Create a mask to clear // bits i through j in n. // EXAMPLE: i = 2, j = 4. // Result should be 11100011. // For simplicity, we'll use // just 8 bits for the example. // will equal sequence of all ls $allOnes = ~0; // ls before position j, then // 0s. left = 11100000 $left= $allOnes << ($j + 1); // l's after position i. // right = 00000011 $right = ((1 << $i) - 1); // All ls, except for 0s between // i and j. mask 11100011 $mask = $left | $right; // Clear bits j through i // then put min there // Clear bits j through i. $n_cleared = $n & $mask; // Move m into correct position. $m_shifted = $m << $i; // OR them, and we're done! return ($n_cleared | $m_shifted); }// Driver Code// in Binary N= 10000000000$n = 1024; // in Binary M= 10011$m = 19; $i = 2;$j = 6;echo updateBits($n, $m, $i, $j);// This code is contributed by Ajit?> |
Javascript
<script>// JavaScript program for implementation of updateBits() // Function to updateBits M insert to N. function updateBits(n,m,i,j) { /* Create a mask to clear bits i through j in n. EXAMPLE: i = 2, j = 4. Result should be 11100011. For simplicity, we'll use just 8 bits for the example. */ let allOnes = ~0; // will equal sequence of all ls // ls before position j, then 0s. left = 11100000 let left= allOnes << (j + 1); // l's after position i. right = 00000011 let right = ((1 << i) - 1); // All ls, except for 0s between i and j. mask 11100011 let mask = left | right; /* Clear bits j through i then put min there */ // Clear bits j through i. let n_cleared = n & mask; // Move m into correct position. let m_shifted = m << i; // OR them, and we're done! return (n_cleared | m_shifted); } // in Binary N= 10000000000 let n = 1024; // in Binary M= 10011 let m = 19; let i = 2, j = 6; document.write(updateBits(n,m,i,j)); // This code is contributed by sravan</script> |
Output
1100
Time Complexity: O(1)
Auxiliary Space: O(1)
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