Binary search using c programming

Question

Given an array of integers which is sorted in ascending order, and an integer target, write a function to search target in the array. If the target exists, then return its index. Otherwise, return -1.

You must write an algorithm with O(log n) runtime complexity.

I solved the above question using the following code:

int search(int* nums, int numsSize, int target){

    int l = 0;
    int r = numsSize -1;
    int result = -1;
    
    while(l <= r)
    {
        int mid = (l + (r-l))/2;
        
        if(nums[mid] == target)
            return result = nums[mid];
        if(nums[mid] < target)
            return l = mid + 1;
        else
            return r = mid - 1;
    }
    return result;
}

But I am still not getting the right answer. Please help.

Answer 1

You are returning the newly calculated l and r value.

if(nums[mid] < target)
    return l = mid + 1;
else
    return r = mid - 1;

You just need to update them and keep searching the element.

if(nums[mid] < target)
    l = mid + 1;
else
    r = mid - 1;

Answer 2

Corrected code:

int search(int* nums, int numsSize, int target){

    int l = 0;
    int r = numsSize -1;
    int result = -1;
    printf("%d %d %d\n", numsSize, target, nums[0]);

    while(l <= r)
    {
        int mid = l + (r - l) / 2;
        
        if(nums[mid] == target)
            return result = mid;
        if(nums[mid] < target)
            l = mid + 1;
        else
            r = mid - 1;
    }
    return result;
}

The errors:

1. You computed wrongly the mid, it's not:

   int mid = (l + (rl))/2;

Correct version is:

int mid = l + (r - l) / 2;

2. You should not use return here because you break the loop.

if(nums[mid] < target)
        return l = mid + 1;
    else
        return r = mid - 1;

    if(nums[mid] == target)
          return result = nums[mid];

Here you should return the mid, the position of the target value, not the value itself.

Answer 3

Try this: It is Java though

 public int search(int[] nums, int target) {
    int l = 0;
    int r = nums.length - 1;
    while(l<=r){
        
        int mid = (l+r)/2;
        if(target == nums[l]){
            return l;
        } else if (target == nums[r]){
            return r;
        } else if(target == nums[mid]){
            return mid;
        }
        else if(target > nums[mid]){
            l = mid;
            r = r - 1;
        }
        else {
            l = l + 1;
            r = mid;
        }
    }
    return -1;
}

Answer 4

solved using python

def search(self, nums, target):
    
    lowerIndex = 0
    higherIndex = len(nums) -1

    while lowerIndex < higherIndex:
        middle = (lowerIndex + higherIndex) // 2
        mid_number = nums[middle]
    
        if mid_number == target:
            return middle
    
        elif mid_number > target:
            lowerIndex = middle + 1
        elif mid_number < target:
            higherIndex = middle -1
    return -1

Answer 5

Im getting TIME LIMIT EXCEEDED