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Java-based LeetCode algorithm problem solutions, regularly updated

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package g2601_2700.s2659_make_array_empty;

// #Hard #Array #Sorting #Greedy #Binary_Search #Ordered_Set #Segment_Tree #Binary_Indexed_Tree
// #2023_09_07_Time_42_ms_(98.92%)_Space_55.3_MB_(95.34%)

import java.util.Arrays;
import java.util.Comparator;

/**
 * 2659 - Make Array Empty\.
 *
 * Hard
 *
 * You are given an integer array `nums` containing **distinct** numbers, and you can perform the following operations **until the array is empty**:
 *
 * *   If the first element has the **smallest** value, remove it
 * *   Otherwise, put the first element at the **end** of the array.
 *
 * Return _an integer denoting the number of operations it takes to make_ `nums` _empty._
 *
 * **Example 1:**
 *
 * **Input:** nums = [3,4,-1]
 *
 * **Output:** 5
 *
 *     Operation   Array
 *     1           [4, -1, 3]
 *     2           [-1, 3, 4]
 *     3           [3, 4]
 *     4           [4]
 *     5           []
 *
 * **Example 2:**
 *
 * **Input:** nums = [1,2,4,3]
 *
 * **Output:** 5
 *
 *     Operation   Array
 *     1           [2, 4, 3]
 *     2           [4, 3]
 *     3           [3, 4]
 *     4           [4]
 *     5           []
 *
 * **Example 3:**
 *
 * **Input:** nums = [1,2,3]
 *
 * **Output:** 3
 *
 *     Operation   Array
 *     1           [2, 3]
 *     2           [3]
 *     3           []
 *
 * **Constraints:**
 *
 * *   1 <= nums.length <= 10⁵
 * *   -10⁹<= nums[i] <= 10⁹
 * *   All values in `nums` are **distinct**.
**/
public class Solution {
    public long countOperationsToEmptyArray(int[] nums) {
        int n = nums.length;
        Integer[] arr = new Integer[n];
        for (int i = 0; i < n; i++) {
            arr[i] = i;
        }
        Arrays.sort(arr, Comparator.comparingInt(i -> nums[i]));
        long ans = n;
        for (int i = 1; i < n; i++) {
            if (arr[i] < arr[i - 1]) {
                ans += n - i;
            }
        }
        return ans;
    }
}