## Set all bits of a number in a range equal to another number

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).
EXAMPLE:
Input: N = 10000000000, M = 10101, i = 2, j = 6
Output: N = 10001010100

My initial thoughts:

1. Create a mask with 1s but only 0s between positions i and j. AND the mask with N to clear the bits between i and j to 0s.
2. Create another mask with 0s but only 1s between positions i and j. AND the mask with M to clear the bits outside i and j to 0s.
3. OR N and M.

My initial codes:

```	public static void setBits(int N, int M, int i, int j) {
for (int digit = i; digit <= j; ++digit)
N = N | M;
}

It does not work…
Solution:
public static int updateBits(int n, int m, int i, int j) {
int max = ~0; /* All 1's */

// 1's through position j, then 0's
int left = max - ((1 << j) - 1);

// 1's after position i
int right = ((1 << i) - 1);

// 1's with 0s between i and j
int mask = left | right;

// Clear i through j, then put m in there
return (n & mask) | (m << i);
}

The idea is the same as mine but my implementation is wrong…

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Related
```

public static int updateBits(int n, int m, int i, int j) {
int max = ~0; /* All 1’s */

// 1’s through position j, then 0’s
int left = max – ((1 << (j+1)) – 1); // here should be j+1 not j.

// 1’s after position i
int right = ((1 << i) – 1);

// 1’s, with 0s between i and j
int mask = left | right;

// Clear i through j, then put m in there
return (n & mask) | (m << i);
}