排序数组中所有可能的 K 长度子序列的中间值的最小和
原文:https://www . geeksforgeeks . org/所有可能的 k 长度排序数组的子序列的最小中值之和/
给定一个排序的数组arr【】由 N 个整数和一个正整数 K (使得 N%K 为 0 )组成，任务是找到大小为 K的所有可能子序列的中值的最小和，使得每个元素只属于一个子序列。
示例:
输入: arr[] = {1，2，3，4，5，6}，K = 2 输出: 6 解释: 考虑大小为K 的子序列为{1，4}、{2，5}和{3，6}。 所有子序列的中值之和为(1 + 2 + 3) = 6，这是最小可能和。
输入: K = 3，arr[] = {3，11，12，22，33，35，38，67，69，71，94，99}，K = 3 输出: 135
天真方法:给定的问题可以通过生成所有可能的 K 大小的排序子序列并打印所有这些子序列的中值作为结果来解决。
时间复杂度:O(2N) 辅助空间: O(1)
有效方法:上述方法也可以通过使用贪婪方法来构建所有子序列来优化。其思想是从数组的开始处选择 K/2 元素，从数组的结束处选择 K/2 元素，这样中间值总是出现在第一部分。按照以下步骤解决问题:
- 初始化一个变量，比如说 res ，它存储中间值的和。
- 初始化一个变量，说 T 为 N/K 存储所需的子序列数，说一个变量 D 为 (K + 1)/2 存储中间值之间的距离。
- 初始化一个变量，说 i 为(D–1)来存储第一个中值的索引以添加到结果中。
- 迭代至 i < N 和T>0T5】的值，执行以下步骤:
- 将arr【I】的值添加到变量 res 中。
- 将 i 的值增加 D 得到下一个中位数的指数。
- 将 T 的值减少 1 。
- 完成上述步骤后，打印 res 的值作为结果。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the minimum sum of
// all the medians of the K sized sorted
// arrays formed from the given array
void sumOfMedians(int arr[], int N,
int K)
{
// Stores the distance between
// the medians
int selectMedian = (K + 1) / 2;
// Stores the number of subsequences
// required
int totalArrays = N / K;
// Stores the resultant sum
int minSum = 0;
// Iterate from start and add
// all the medians
int i = selectMedian - 1;
while (i < N and totalArrays != 0) {
// Add the value of arr[i]
// to the variable minsum
minSum = minSum + arr[i];
// Increment i by select the
// median to get the next
// median index
i = i + selectMedian;
// Decrement the value of
// totalArrays by 1
totalArrays--;
}
// Print the resultant minimum sum
cout << minSum;
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6 };
int N = sizeof(arr) / sizeof(int);
int K = 2;
sumOfMedians(arr, N, K);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
public class GFG {
// Function to find the minimum sum of
// all the medians of the K sized sorted
// arrays formed from the given array
static void sumOfMedians(int arr[], int N, int K)
{
// Stores the distance between
// the medians
int selectMedian = (K + 1) / 2;
// Stores the number of subsequences
// required
int totalArrays = N / K;
// Stores the resultant sum
int minSum = 0;
// Iterate from start and add
// all the medians
int i = selectMedian - 1;
while (i < N && totalArrays != 0) {
// Add the value of arr[i]
// to the variable minsum
minSum = minSum + arr[i];
// Increment i by select the
// median to get the next
// median index
i = i + selectMedian;
// Decrement the value of
// totalArrays by 1
totalArrays--;
}
// Print the resultant minimum sum
System.out.println(minSum);
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 1, 2, 3, 4, 5, 6 };
int N = arr.length;
int K = 2;
sumOfMedians(arr, N, K);
}
}
// This code is contributed by Kingash.
Python 3
# Python3 program for the above approach
# Function to find the minimum sum of
# all the medians of the K sized sorted
# arrays formed from the given array
def sumOfMedians(arr, N, K):
# Stores the distance between
# the medians
selectMedian = (K + 1) // 2
# Stores the number of subsequences
# required
totalArrays = N // K
# Stores the resultant sum
minSum = 0
# Iterate from start and add
# all the medians
i = selectMedian - 1
while (i < N and totalArrays != 0):
# Add the value of arr[i]
# to the variable minsum
minSum = minSum + arr[i]
# Increment i by select the
# median to get the next
# median index
i = i + selectMedian
# Decrement the value of
# totalArrays by 1
totalArrays -= 1
# Print the resultant minimum sum
print(minSum)
# Driver Code
if __name__ == '__main__':
arr = [ 1, 2, 3, 4, 5, 6 ]
N = len(arr)
K = 2
sumOfMedians(arr, N, K)
# This code is contributed by nirajgsuain5
C
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the minimum sum of
// all the medians of the K sized sorted
// arrays formed from the given array
static void sumOfMedians(int[] arr, int N, int K)
{
// Stores the distance between
// the medians
int selectMedian = (K + 1) / 2;
// Stores the number of subsequences
// required
int totalArrays = N / K;
// Stores the resultant sum
int minSum = 0;
// Iterate from start and add
// all the medians
int i = selectMedian - 1;
while (i < N && totalArrays != 0) {
// Add the value of arr[i]
// to the variable minsum
minSum = minSum + arr[i];
// Increment i by select the
// median to get the next
// median index
i = i + selectMedian;
// Decrement the value of
// totalArrays by 1
totalArrays--;
}
// Print the resultant minimum sum
Console.WriteLine(minSum);
}
// Driver Code
public static void Main()
{
int[] arr = { 1, 2, 3, 4, 5, 6 };
int N = arr.Length;
int K = 2;
sumOfMedians(arr, N, K);
}
}
// This code is contributed by code_hunt.
java 描述语言
<script>
// JavaScript program for the above approach
// Function to find the minimum sum of
// all the medians of the K sized sorted
// arrays formed from the given array
function sumOfMedians(arr, N, K)
{
// Stores the distance between
// the medians
let selectMedian = Math.floor((K + 1) / 2);
// Stores the number of subsequences
// required
let totalArrays = Math.floor(N / K);
// Stores the resultant sum
let minSum = 0;
// Iterate from start and add
// all the medians
let i = selectMedian - 1;
while (i < N && totalArrays != 0) {
// Add the value of arr[i]
// to the variable minsum
minSum = minSum + arr[i];
// Increment i by select the
// median to get the next
// median index
i = i + selectMedian;
// Decrement the value of
// totalArrays by 1
totalArrays--;
}
// Print the resultant minimum sum
document.write(minSum);
}
// Driver Code
let arr = [ 1, 2, 3, 4, 5, 6 ];
let N = arr.length;
let K = 2;
sumOfMedians(arr, N, K);
</script>
Output:
6
时间复杂度:O(N) T5辅助空间:** O(1)
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