• Home
• com.github.javadev
• leetcode-in-java21
• 1.26
• source code
• Solution.java

×

Project download

Many resources are needed to download a project. Please understand that we have to compensate our server costs. Thank you in advance.
Project price only 1 $

You can buy this project and download/modify it how often you want.

g2601_2700.s2681_power_of_heroes.Solution Maven / Gradle / Ivy

package g2601_2700.s2681_power_of_heroes;

// #Hard #Array #Math #Sorting #Prefix_Sum #2023_09_11_Time_13_ms_(88.24%)_Space_54.4_MB_(55.29%)

import java.util.Arrays;

/**
 * 2681 - Power of Heroes\.
 *
 * Hard
 *
 * You are given a **0-indexed** integer array `nums` representing the strength of some heroes. The **power** of a group of heroes is defined as follows:
 *
 * *   Let i0, i1, ... ,ik be the indices of the heroes in a group. Then, the power of this group is max(nums[i0], nums[i1], ... ,nums[ik])2 * min(nums[i0], nums[i1], ... ,nums[ik]).
 *
 * Return _the sum of the **power** of all **non-empty** groups of heroes possible._ Since the sum could be very large, return it **modulo** 109 + 7.
 *
 * **Example 1:**
 *
 * **Input:** nums = [2,1,4]
 *
 * **Output:** 141
 *
 * **Explanation:** 
 *
 * 1st group: [2] has power = 22 \* 2 = 8. 
 *
 * 2nd group: [1] has power = 12 \* 1 = 1. 
 *
 * 3rd group: [4] has power = 42 \* 4 = 64. 
 *
 * 4th group: [2,1] has power = 22 \* 1 = 4. 
 *
 * 5th group: [2,4] has power = 42 \* 2 = 32. 
 *
 * 6th group: [1,4] has power = 42 \* 1 = 16. 
 *
 * 7th group: [2,1,4] has power = 42 \* 1 = 16. 
 *
 * The sum of powers of all groups is 8 + 1 + 64 + 4 + 32 + 16 + 16 = 141.
 *
 * **Example 2:**
 *
 * **Input:** nums = [1,1,1]
 *
 * **Output:** 7
 *
 * **Explanation:** A total of 7 groups are possible, and the power of each group will be 1. Therefore, the sum of the powers of all groups is 7.
 *
 * **Constraints:**
 *
 * *   1 <= nums.length <= 105
 * *   1 <= nums[i] <= 109
**/
public class Solution {
    public int sumOfPower(int[] nums) {
        Arrays.sort(nums);
        long sumOfMin = 0L;
        long res = 0L;
        for (int num : nums) {
            long mod = 1_000_000_007L;
            long max = (long) num * num % mod;
            long mySumOfMin = (sumOfMin + num) % mod;
            res = (res + max * mySumOfMin) % mod;
            sumOfMin = (sumOfMin + mySumOfMin) % mod;
        }
        return (int) res;
    }
}