Problem
Given an integer array nums
, your goal is to make all elements in nums
equal. To complete one operation, follow these steps:
Find the largest value in
nums
. Let its index bei
(0-indexed) and its value belargest
. If there are multiple elements with the largest value, pick the smallesti
.Find the next largest value in
nums
strictly smaller thanlargest
. Let its value benextLargest
.Reduce
nums[i]
tonextLargest
.
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:
1 <= nums.length <= 5 * 104
1 <= nums[i] <= 5 * 104
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:
- Time complexity : O(n * log(n)).
- Space complexity : O(n).