最小化插入使每对连续数组元素之和至多 K

给定一个由N 个正整数组成的数组arr[]和一个整数K ，任务是找到任何正整数的最小插入次数，使得每个相邻元素的总和最多为 K 。如果不可能，则打印“-1” 。

例子：

Input: arr[] = {1, 2, 3, 4, 5}, K = 6
Output: 2
Explanation:
Following insertions are required:
Operation 1: Insert 1 between indices 2 and 3. Therefore, the array modifies to {1, 2, 3, 1, 4, 5}.
Operation 2: Insert 1 between index 4 and 5. Therefore, the array modifies to {1, 2, 3, 1, 4, 1, 5}. Therefore, the minimum number of insertions required is 2.

Input: arr[] = {4, 5, 6, 7, 7, 8}, K = 8
Output: -1

方法：这个想法基于这样一个事实，如果元素本身不等于或大于K ，则在总和超过K的元素之间插入1会使连续元素的总和小于K 。请按照以下步骤解决给定的问题：

• 初始化三个变量，比如res = 0 、可能 = 1和last = 0来存储最小插入次数，检查是否有可能使所有连续对的总和最多为 K ，并存储前一个编号分别为当前元素。
• 遍历数组arr[]并执行以下步骤：
  • 如果arr[i]的值至少为K ，则不可能使所有连续对的总和至多为 K 。
  • 如果last和arr[i]的总和大于K ，则将res增加1并且 分配last = arr[i] 。
• 完成上述步骤后，如果可能的值为1 ，则将res的值打印为所需的最小插入次数。否则，打印“-1” 。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to count minimum number of
// insertions required to make sum of
// every pair of adjacent elements at most K
void minimumInsertions(int arr[],
                       int N, int K)
{
    // Stores if it is possible to
    // make sum of each pair of
    // adjacent elements at most K
    bool possible = 1;
 
    // Stores the count of insertions
    int res = 0;
 
    // Stores the previous
    // value to index i
    int last = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // If arr[i] is greater
        // than or equal to K
        if (arr[i] >= K) {
 
            // Mark possible 0
            possible = 0;
            break;
        }
 
        // If last + arr[i]
        // is greater than K
        if (last + arr[i] > K)
 
            // Increment res by 1
            res++;
 
        // Assign arr[i] to last
        last = arr[i];
    }
 
    // If possible to make the sum of
    // pairs of adjacent elements at most K
    if (possible) {
        cout << res;
    }
 
    // Otherwise print "-1"
    else {
 
        cout << "-1";
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int K = 6;
    int N = sizeof(arr) / sizeof(arr[0]);
 
    minimumInsertions(arr, N, K);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG
{
 
  // Function to count minimum number of
  // insertions required to make sum of
  // every pair of adjacent elements at most K
  static void minimumInsertions(int arr[],
                                int N, int K)
  {
 
    // Stores if it is possible to
    // make sum of each pair of
    // adjacent elements at most K
    boolean possible = true;
 
    // Stores the count of insertions
    int res = 0;
 
    // Stores the previous
    // value to index i
    int last = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
      // If arr[i] is greater
      // than or equal to K
      if (arr[i] >= K) {
 
        // Mark possible 0
        possible = false;
        break;
      }
 
      // If last + arr[i]
      // is greater than K
      if (last + arr[i] > K)
 
        // Increment res by 1
        res++;
 
      // Assign arr[i] to last
      last = arr[i];
    }
 
    // If possible to make the sum of
    // pairs of adjacent elements at most K
    if (possible) {
      System.out.print(res);
    }
 
    // Otherwise print "-1"
    else {
 
      System.out.print("-1");
    }
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int arr[] = { 1, 2, 3, 4, 5 };
    int K = 6;
    int N = arr.length;
 
    minimumInsertions(arr, N, K);
 
  }
}
 
// This code is contributed by shikhasingrajput

Python3

# Python 3 program for the above approach
 
# Function to count minimum number of
# insertions required to make sum of
# every pair of adjacent elements at most K
def minimumInsertions(arr, N, K):
 
    # Stores if it is possible to
    # make sum of each pair of
    # adjacent elements at most K
    possible = 1
 
    # Stores the count of insertions
    res = 0
 
    # Stores the previous
    # value to index i
    last = 0
 
    # Traverse the array
    for i in range(N):
 
        # If arr[i] is greater
        # than or equal to K
        if (arr[i] >= K):
 
            # Mark possible 0
            possible = 0
            break
 
        # If last + arr[i]
        # is greater than K
        if (last + arr[i] > K):
 
            # Increment res by 1
            res += 1
 
        # Assign arr[i] to last
        last = arr[i]
 
    # If possible to make the sum of
    # pairs of adjacent elements at most K
    if (possible):
        print(res)
 
    # Otherwise print "-1"
    else:
        print("-1")
 
# Driver Code
if __name__ == "__main__":
 
    arr = [1, 2, 3, 4, 5]
    K = 6
    N = len(arr)
 
    minimumInsertions(arr, N, K)
 
    # This code is contributed by ukasp.

C#

// C# program for the above approach
using System;
 
class GFG{
 
  // Function to count minimum number of
  // insertions required to make sum of
  // every pair of adjacent elements at most K
  static void minimumInsertions(int[] arr,
                                int N, int K)
  {
 
    // Stores if it is possible to
    // make sum of each pair of
    // adjacent elements at most K
    bool possible = true;
 
    // Stores the count of insertions
    int res = 0;
 
    // Stores the previous
    // value to index i
    int last = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
      // If arr[i] is greater
      // than or equal to K
      if (arr[i] >= K) {
 
        // Mark possible 0
        possible = false;
        break;
      }
 
      // If last + arr[i]
      // is greater than K
      if (last + arr[i] > K)
 
        // Increment res by 1
        res++;
 
      // Assign arr[i] to last
      last = arr[i];
    }
 
    // If possible to make the sum of
    // pairs of adjacent elements at most K
    if (possible) {
      Console.Write(res);
    }
 
    // Otherwise print "-1"
    else {
 
      Console.Write("-1");
    }
  }
 
  // Driver code
  static void Main()
  {
 
    int[] arr = { 1, 2, 3, 4, 5 };
    int K = 6;
    int N = arr.Length;
 
    minimumInsertions(arr, N, K);
  }
}
 
// This code is contributed by sanjoy_62.

Javascript

 
// Javascript program for the above approach
 
    // Function to count minimum number of
    // insertions required to make sum of
    // every pair of adjacent elements at most K
    function minimumInsertions(arr , N , K) {
 
        // Stores if it is possible to
        // make sum of each pair of
        // adjacent elements at most K
        var possible = true;
 
        // Stores the count of insertions
        var res = 0;
 
        // Stores the previous
        // value to index i
        var last = 0;
 
        // Traverse the array
        for (i = 0; i < N; i++) {
 
            // If arr[i] is greater
            // than or equal to K
            if (arr[i] >= K) {
 
                // Mark possible 0
                possible = false;
                break;
            }
 
            // If last + arr[i]
            // is greater than K
            if (last + arr[i] > K)
 
                // Increment res by 1
                res++;
 
            // Assign arr[i] to last
            last = arr[i];
        }
 
        // If possible to make the sum of
        // pairs of adjacent elements at most K
        if (possible) {
            document.write(res);
        }
 
        // Otherwise prvar "-1"
        else {
 
            document.write("-1");
        }
    }
 
    // Driver Code
     
        var arr = [ 1, 2, 3, 4, 5 ];
        var K = 6;
        var N = arr.length;
 
        minimumInsertions(arr, N, K);
 
 
// This code contributed by umadevi9616
 

输出：

2

时间复杂度： O(N)
辅助空间： O(1)

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