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Java-based LeetCode algorithm problem solutions, regularly updated

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package g2501_2600.s2592_maximize_greatness_of_an_array;

// #Medium #Array #Sorting #Greedy #Two_Pointers
// #2023_08_23_Time_8_ms_(100.00%)_Space_57.2_MB_(41.49%)

import java.util.Arrays;

/**
 * 2592 - Maximize Greatness of an Array\.
 *
 * Medium
 *
 * You are given a 0-indexed integer array `nums`. You are allowed to permute `nums` into a new array `perm` of your choosing.
 *
 * We define the **greatness** of `nums` be the number of indices `0 <= i < nums.length` for which `perm[i] > nums[i]`.
 *
 * Return _the **maximum** possible greatness you can achieve after permuting_ `nums`.
 *
 * **Example 1:**
 *
 * **Input:** nums = [1,3,5,2,1,3,1]
 *
 * **Output:** 4
 *
 * **Explanation:** One of the optimal rearrangements is perm = [2,5,1,3,3,1,1]. At indices = 0, 1, 3, and 4, perm[i] > nums[i]. Hence, we return 4.
 *
 * **Example 2:**
 *
 * **Input:** nums = [1,2,3,4]
 *
 * **Output:** 3
 *
 * **Explanation:** We can prove the optimal perm is [2,3,4,1]. At indices = 0, 1, and 2, perm[i] > nums[i]. Hence, we return 3.
 *
 * **Constraints:**
 *
 * *   1 <= nums.length <= 10⁵
 * *   0 <= nums[i] <= 10⁹
**/
public class Solution {
    public int maximizeGreatness(int[] nums) {
        Arrays.sort(nums);
        int perm = 0;
        for (int num : nums) {
            if (nums[perm] < num) {
                perm++;
            }
        }
        return perm;
    }
}