Using Binary Search with sorted Array with duplicates [duplicate]

Answer 1

Another result for log(n) binary search for leftmost target and rightmost target. This is in C++, but I think it is quite readable.

The idea is that we always end up when left = right + 1. So, to find leftmost target, if we can move right to rightmost number which is less than target, left will be at the leftmost target.

For leftmost target:

int binary_search(vector& nums, int target){
    int n = nums.size();
    int left = 0, right = n - 1;

    // carry right to the greatest number which is less than target.
    while(left <= right){
        int mid = (left + right) / 2;
        if(nums[mid] < target)
            left = mid + 1;
        else
            right = mid - 1;
    }
    // when we are here, right is at the index of greatest number
    // which is less than target and since left is at the next, 
    // it is at the first target's index
    return left;
}

For the rightmost target, the idea is very similar:

int binary_search(vector& nums, int target){
    while(left <= right){
        int mid = (left + right) / 2;
        // carry left to the smallest number which is greater than target.
        if(nums[mid] <= target)
            left = mid + 1;
        else
            right = mid - 1;
    }
    // when we are here, left is at the index of smallest number
    // which is greater than target and since right is at the next, 
    // it is at the first target's index
    return right;
}