[Programming Problem] Find Minimum in Rotated Sorted Array II

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

  • [4,5,6,7,0,1,4] if it was rotated 4 times.
  • [0,1,4,4,5,6,7] 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 that may contain duplicates, return the minimum element of this array.

You must decrease the overall operation steps as much as possible.

Example 1:
Input: nums = [1,3,5]
Output: 1

Example 2:
Input: nums = [2,2,2,0,1]
Output: 0

[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. Because of duplicates, we might be in a situation where there’s no one clear winner. In that case, move in both directions.
    if ( (a < c && a < d) || (b < c && b < d) ) // go left
    else if ((c > a && c > b) || (d > a && d > b)) // go right
    else // go both directions.
  • 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) {    
    if (i === j) return nums[i];
 
    let mid = i + Math.floor((j-i)/2);
    const a = nums[i];
    const b = nums[mid];
    const c = nums[mid + 1];
    const d = nums[j];
 
    let left = Number.MAX_VALUE;
    let right = Number.MAX_VALUE;
    if ((a < c && a < d) || (b < c && b < d)) {
        // go left
        left = findMinRecursive(nums, i, mid);
    } else if ((c > a && c > b) || (d > a && d > b)) {
        right = findMinRecursive(nums, mid+1, j);
    } else {
        // go both
        left = findMinRecursive(nums, i, mid);
        right = findMinRecursive(nums, mid+1, j);
    }
 
    return Math.min(left, right);
};

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