# 5.1. Insertion > You're given 2 32-bit numbers, N and M, and two bit positions, i and j. Write a method to insert M into N such that M starts at bit j and ends at bit i. Assume that bits j through i have enough space to fit all of M. If M=10011, you can assume there are at least 5 bits between j and i. Example: * Input: N = 10000000000, M=10011, i=2, j=6 * Output: N = 10001001100 ## Initial idea I guess we have to *update* bits starting at j (from N). Reusing the update function from README.md. Probably wrong but idea is to update all bits of M iteratively. O(M). ```c++ int insertBit(int N, int M, int j) { String newM = Integer.toBinaryString(M); for(size_t it = 0; it < newM.length; it++){ int value = int(it) ? 1 : 0; int mask = ~(1 << j); N += int((num & mask) | (value << j)); } return N; } ``` Another idea could be to slice up N into [:j] and [i+1:]. and then slice_1 + M + slice_2. But this would have to be done in string mode, I think the focus here is doing it with int? ## Hint 1 > Focus first on clearing the appropriate bits This is included in the update function I used before. ## Hint 2 > To clear bits, create a 'bit mask' that looks like a series of 1s, then 0s, then 1s. `int mask = (1 << X) + (0 << (j-i+1)) + (1 << i);` How to get size of N? it says 32 bits but in the example there were 10. `N = int(N & mask) | (M << j)` or something. ## Hint 3 > To create the mask, create a bit mask for the left side, and another one for the right side. Then merge I think I've done that already. ## Solution 1. Clear bits j through i in N: mask 2. Shift M so that it lines up with bits j through i 3. Merge M and N ```java int updateBits(int n, int m, int i, int j) { if (i > j || i < 0 || j >= 32) return 0; // forgot to add this /* Create mask to clear bits i through j in n. if i=2, and j=4, it should look like 11100011. For simplicity we use 8 bits for the example */ int allOnes = ~0; // 11100000 int left = j < 31 ? (allOnes << (j+1)) : 0; // 00000011 int right = ((1 << i) - 1); int mask = left | right; // clear bits j through i and then put m in there int n_cleared = n & mask; int m_shifted = m << i; return n_cleared | m_shifted; } ```