1887. Reduction Operations to Make the Array Elements Equal

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    Problem

    Given an integer array nums, your goal is to make all elements in nums equal. To complete one operation, follow these steps:

    Return **the number of operations to make all elements in *nums* equal**.

      Example 1:

    Input: nums = [5,1,3]
    Output: 3
    Explanation: It takes 3 operations to make all elements in nums equal:
    1. largest = 5 at index 0. nextLargest = 3. Reduce nums[0] to 3. nums = [3,1,3].
    2. largest = 3 at index 0. nextLargest = 1. Reduce nums[0] to 1. nums = [1,1,3].
    3. largest = 3 at index 2. nextLargest = 1. Reduce nums[2] to 1. nums = [1,1,1].
    

    Example 2:

    Input: nums = [1,1,1]
    Output: 0
    Explanation: All elements in nums are already equal.
    

    Example 3:

    Input: nums = [1,1,2,2,3]
    Output: 4
    Explanation: It takes 4 operations to make all elements in nums equal:
    1. largest = 3 at index 4. nextLargest = 2. Reduce nums[4] to 2. nums = [1,1,2,2,2].
    2. largest = 2 at index 2. nextLargest = 1. Reduce nums[2] to 1. nums = [1,1,1,2,2].
    3. largest = 2 at index 3. nextLargest = 1. Reduce nums[3] to 1. nums = [1,1,1,1,2].
    4. largest = 2 at index 4. nextLargest = 1. Reduce nums[4] to 1. nums = [1,1,1,1,1].
    

      Constraints:

    Solution

    /**
     * @param {number[]} nums
     * @return {number}
     */
    var reductionOperations = function(nums) {
        nums.sort((a, b) => a - b);
        var uniqueNums = [];
        for (var i = 0; i < nums.length; i++) {
            if (nums[i] === nums[i - 1]) {
                uniqueNums[uniqueNums.length - 1]++;
            } else {
                uniqueNums.push(1);
            }
        }
        var res = 0;
        while (uniqueNums.length > 1) {
            var last = uniqueNums.pop();
            res += last;
            uniqueNums[uniqueNums.length - 1] += last;
        }
        return res;
    };
    

    Explain:

    nope.

    Complexity: