Find Minimum in Rotated Sorted Array

[kokocan12]

Problem

When an array is given, that is sorted in ascending order.
But it is rotated 1~n times.

When array is [0,1,2,4,5,6,7]
[4,5,6,7,0,1,2] if it was rotated 4 times.
[0,1,2,4,5,6,7] if it was rotated 7 times.

Approach

First, find out it is rotated or not.
If it rotated n times that is its length, it is same as default.

Then, find out which part the initial value exist by comparing mid value of array with start value of array.
It can be in left part or right part.

The code is below.

/**
 *  [0,1,2,3,4,5,6,7] not rotated
 *  [7,0,1,2,3,4,5,6] rotated && nums[mid] < nums[0] Then MIN is in the LEFT-SIDE
 *  [6,7,0,1,2,3,4,5] rotated && nums[mid] < nums[0] Then MIN is in the LEFT-SIDE
 *  [5,6,7,0,1,2,3,4] rotated && nums[mid] < nums[0] Then MIN is in the LEFT-SIDE
 *  [4,5,6,7,0,1,2,3] rotated && nums[mid] >= nums[0] Then MIN is in the RIGHT-SIDE
 *  [3,4,5,6,7,0,1,2] rotated && nums[mid] >= nums[0] Then MIN is in the RIGHT-SIDE
 *  [2,3,4,5,6,7,0,1] rotated && nums[mid] >= nums[0] Then MIN is in the RIGHT-SIDE
 *  [1,2,3,4,5,6,7,0] rotated && nums[mid] >= nums[0] Then MIN is in the RIGHT-SIDE
 *  [0,1,2,3,4,5,6,7] not rotated
 *
 *  [3,4,5,6,7,0,1,2] rotated,  nums[mid] >= nums[0] MIN is in the RIGHT-SIDE
 *  [7,0,1,2] rotated, nums[mid] < nums[0] MIN is in the LEFT-SIDE
 *  [7,0] rotated, nums[mid] >= nums[0] MIN is in the RIGHT-SIDE
 *  [0] MIN detected.
 *
 *  [4,5,6,7,0,1,2,3] rotated  nums[mid] >= nums[0] MIN is in the RIGHT-SIDE
 *  [0,1,2,3] is not rotated, return nums[0];
 *
 *  [2,3,4,5,6,7,0,1] rotated  nums[mid] >= nums[0] MIN is in the RIGHT-SIDE
 *  [6,7,0,1] rotated, nums[mid] >= nums[0] MIN is in the RIGHT-SIDE
 *  [0,1] not rotated, return nums[0];
 */

function findMinValue(nums, s, e) {
    // If not rotated
    if(nums[s] < nums[e] || s===e) return nums[s];
    
    // If rotated.
    const mid = (s + e) >> 1;
    
    // Min value is in the left-side.
    if(nums[mid] < nums[s]) {
        return findMinValue(nums, s, mid);
    } 
    // Min value is in the right-side.
    else {
        return findMinValue(nums, mid+1, e);
    }
}