Below is a recursive version of binarySearch: int binarySearch(int nums[], int low, int high, int target) { if (low > high) return -1; int mid = (low + high)/2; if (target == nums[mid]) return mid; else if (target < nums[mid]) return binarySearch(nums, low, mid - 1, target); else return binarySearch(nums, mid + 1, high, target); } Code Analysis () If we change the first line of code in the function to be “if (low >= high) return -1;”, and we have an array defined as “int nums[] = {2, 3, 5, 7, 8, 10};”, answer the questions below if we called this function with “int index = binarySearch(nums, 0, nums.length - 1, 3);” For each iteration of binarySearch: 3) If the base case(s) are false, which recursive function call will be made
Below is a recursive version of binarySearch:
int binarySearch(int nums[], int low, int high, int target)
{
if (low > high)
return -1;
int mid = (low + high)/2;
if (target == nums[mid])
return mid;
else if (target < nums[mid])
return binarySearch(nums, low, mid - 1, target);
else
return binarySearch(nums, mid + 1, high, target);
}
Code Analysis ()
If we change the first line of code in the function to be “if (low >= high) return -1;”, and we have an array defined as “int nums[] = {2, 3, 5, 7, 8, 10};”, answer the questions below if we called this function with “int index = binarySearch(nums, 0, nums.length - 1, 3);”
For each iteration of binarySearch:
3) If the base case(s) are false, which recursive function call will be made?
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