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# Set bits in N equals to M in the given range.

• Difficulty Level : Medium
• Last Updated : 29 Jun, 2021

You are given two 32-bit numbers, N and M, and two bit positions, i and j. Write a method to set all bits between i and j in N equal to M (e.g., M becomes a substring of N located at i and starting at j).
Examples :

```Input : N = 1, M = 2, i = 2, j = 4
Output: 9
N = 00000001(Considering 8 bits only)
M = 10 (Binary of 2) For more indexes,
Now set 3 bits from ith index to j in
the N as in the M.
Bits:-    0 0 0 (0  1  0) 0 1 = 9
Indexes:- 7 6 5  4  3  2  1 0
From index 2 to 4, bits are set according
to the M.```

A simple solution is to traverse all bits in N from 0 to 31 and set the bits equals to M in the range from i to j.
An efficient solution is to do following steps.

1. Set all the bits after j in a number.
2. Set all the bits before i in a number.
3. Then perform Bitwise Or on both then we get the number with all the bits set except from i to j.
4. Perform Bitwise And with the given N as to set the bits according to the N.
5. Then shift M into the correct position i.e. in the range of i to j.
6. And at the last perform Bitwise Or on (Shifted M and the N modified in 4th step).
7. The result will be N with M as substring from ith to jth bits

## C++

 `// C++ program for above implementation``#include ``using` `namespace` `std;` `// Function to set the bits``int` `setBits(``int` `n, ``int` `m, ``int` `i, ``int` `j)``{``    ``// number with all 1's``    ``int` `allOnes = ~0;` `    ``// Set all the bits in the left of j``    ``int` `left = allOnes << (j + 1);` `    ``// Set all the bits in the right of j``    ``int` `right = ((1 << i) - 1);` `    ``// Do Bitwsie OR to get all the bits``    ``// set except in the range from i to j``    ``int` `mask = left | right;` `    ``// clear bits j through i``    ``int` `masked_n = n & mask;` `    ``// move m into the correct position``    ``int` `m_shifted = m << i;` `    ``// return the Bitwise OR of masked_n``    ``// and shifted_m``    ``return` `(masked_n | m_shifted);``}` `// Drivers program``int` `main()``{``    ``int` `n = 2, m = 4;``    ``int` `i = 2, j = 4;``    ``cout << setBits(n, m, i, j);``    ``return` `0;``}`

## Java

 `// Java Program``public` `class` `GFG``{``    ``// Function to set the bits``    ``static` `int` `setBits(``int` `n, ``int` `m, ``int`  `i, ``int` `j)``    ``{``        ` `        ``// number with all 1's``        ``int`  `allOnes = ~``0``;``        ` `        ``// Set all the bits in the left of j``        ``int` `left = allOnes << (j + ``1``);``        ` `        ``// Set all the bits in the right of j``        ``int` `right = ((``1` `<< i) - ``1``);``        ` `        ``// Do Bitwise OR to get all the bits``        ``// set except in the range from i to j``        ``int` `mask = left | right;``        ` `        ``// clear bits j through i``        ``int` `masked_n = n & mask;``        ` `        ``// move m into the correct position``        ``int` `m_shifted = m << i;``        ` `        ``// return the Bitwise OR of masked_n``        ``// and shifted_m``        ``return` `(masked_n | m_shifted);``    ``}``    ` `    ``// Driver Program to test above function``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `n = ``2``, m = ``4``;``        ``int` `i = ``2``, j = ``4``;``        ``System.out.println(setBits(n, m, i, j));``    ``}``}``// This code is contributed by Sumit Ghosh`

## Python

 `# Python program for above implementation` `# Function to set the bits``def` `setBits(n, m, i, j):` `    ``# number with all 1's``    ``allOnes ``=` `not` `0`` ` `    ``# Set all the bits in the left of j``    ``left ``=` `allOnes << (j ``+` `1``)`` ` `    ``# Set all the bits in the right of j``    ``right ``=` `((``1` `<< i) ``-` `1``)`` ` `    ``# Do Bitwsie OR to get all the bits``    ``# set except in the range from i to j``    ``mask ``=` `left | right`` ` `    ``# clear bits j through i``    ``masked_n ``=` `n & mask`` ` `    ``# move m into the correct position``    ``m_shifted ``=` `m << i`` ` `    ``# return the Bitwise OR of masked_n``    ``# and shifted_m``    ``return` `(masked_n | m_shifted)`` ` `# Drivers program``n, m ``=` `2``, ``4``i, j ``=` `2``, ``4``print` `setBits(n, m, i, j)` `# This code is submitted by Sachin Bisht`

## C#

 `// C# Program for above implementation``using` `System;` `public` `class` `GFG {``    ` `    ``// Function to set the bits``    ``static` `int` `setBits(``int` `n, ``int` `m, ``int`  `i, ``int` `j)``    ``{``         ` `        ``// number with all 1's``        ``int`  `allOnes = ~0;``         ` `        ``// Set all the bits in the left of j``        ``int` `left = allOnes << (j + 1);``         ` `        ``// Set all the bits in the right of j``        ``int` `right = ((1 << i) - 1);``         ` `        ``// Do Bitwise OR to get all the bits``        ``// set except in the range from i to j``        ``int` `mask = left | right;``         ` `        ``// clear bits j through i``        ``int` `masked_n = n & mask;``         ` `        ``// move m into the correct position``        ``int` `m_shifted = m << i;``         ` `        ``// return the Bitwise OR of masked_n``        ``// and shifted_m``        ``return` `(masked_n | m_shifted);``    ``}``     ` `    ``// Driver Program to test above function``    ``public` `static` `void` `Main()``    ``{``        ``int` `n = 2, m = 4;``        ``int` `i = 2, j = 4;``        ` `        ``Console.WriteLine(setBits(n, m, i, j));``    ``}``}` `// This code is contributed by Anant Agarwal.`

## PHP

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## Javascript

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Output :

`18`

Reference:
https://www.careercup.com/question?id=8863294
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