[Programming Problem] Find Minimum in Rotated Sorted Array

Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:

  • [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.

Notice that rotating an array [a[0], a[1], a[2], …, a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], …, a[n-2]].

Given the sorted rotated array nums of unique elements, return the minimum element of this array.

You must write an algorithm that runs in O(log n) time.

Example 1:
Input: nums = [3,4,5,1,2]
Output: 1
Explanation: The original array was [1,2,3,4,5] rotated 3 times.

Example 2:
Input: nums = [4,5,6,7,0,1,2]
Output: 0
Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.

Example 3:
Input: nums = [11,13,15,17]
Output: 11
Explanation: The original array was [11,13,15,17] and it was rotated 4 times.

[Problem Link]

We use a `modified` binary search to find the minimum element:-

  • At every point, compare (‘i’, ‘mid’) with (‘mid+1’, ‘j’). Whichever section has the lower value, let’s move that direction.
    if ( (a < c && a < d) || (b < c && b < d) ) // go left
    else // go right
  • Return the last number remaining in the recursion (when i === j)
/**
 * @param {number[]} nums
 * @return {number}
 */
var findMin = function(nums) {
    return findMinRecursive(nums, 0, nums.length-1)
};
 
var findMinRecursive = function(nums, i, j) {
 
    // one or two digits
    if ( i === j ) return nums[i];
 
    let mid = i + Math.floor((j-i)/2);
    let a = nums[i];
    let b = nums[mid];
    let c = nums[mid+1];
    let d = nums[j];
 
    if ( (a < c && a < d) || (b < c && b < d) ) {
        // go left
        return findMinRecursive(nums, i, mid);
    } else {
        // go right
        return findMinRecursive(nums, mid+1, j);
    }
 
}

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