需要删除一对数组元素的打印索引，才能将数组拆分为3个相等和的子数组

给定一个由N个整数组成的数组arr [] ，任务是打印两个需要删除的数组元素的索引，以便可以将给定的数组拆分为三个相等总和的子数组。如果无法这样做，请打印“ -1” 。

例子：

Input: arr[] = {2, 5, 12, 7, 19, 4, 3}
Output: 2 4
Explanation:
Removing arr[2] and arr[4] modifies arr[] to {2, 5, 7, 4, 3}.
Sum of subarray {arr[0], arr[1]} = 7.
arr[2] = 7.
Sum of subarray {arr[3], arr[4]} = 7. 

Input: arr[] = {2, 1, 13, 5, 14}
Output: -1

天真的方法：最简单的方法是生成所有可能的数组元素对，并针对每一对，检查去除这些对是否可以从给定的数组中生成三个相等的子数组。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to check if array can
// be split into three equal sum
// subarrays by removing two elements
void findSplit(int arr[], int N)
{
    for (int l = 1; l <= N - 4; l++) {
 
        for (int r = l + 2; r <= N - 2; r++) {
 
            // Stores sum of all three subarrays
            int lsum = 0, rsum = 0, msum = 0;
 
            // Sum of left subarray
            for (int i = 0; i <= l - 1; i++) {
                lsum += arr[i];
            }
 
            // Sum of middle subarray
            for (int i = l + 1; i <= r - 1; i++) {
                msum += arr[i];
            }
 
            // Sum of right subarray
            for (int i = r + 1; i < N; i++) {
                rsum += arr[i];
            }
 
            // Check if sum of subarrays are equal
            if (lsum == rsum && rsum == msum) {
 
                // Print the possible pair
                cout << l << " " << r << endl;
                return;
            }
        }
    }
 
    // If no pair exists, print -1
    cout << -1 << endl;
}
 
// Driver code
int main()
{
    // Given array
    int arr[] = { 2, 5, 12, 7, 19, 4, 3 };
 
    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    findSplit(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG
{
 
// Function to check if array can
// be split into three equal sum
// subarrays by removing two elements
static void findSplit(int arr[], int N)
{
    for (int l = 1; l <= N - 4; l++)
    {
        for (int r = l + 2; r <= N - 2; r++)
        {
 
            // Stores sum of all three subarrays
            int lsum = 0, rsum = 0, msum = 0;
 
            // Sum of left subarray
            for (int i = 0; i <= l - 1; i++) {
                lsum += arr[i];
            }
 
            // Sum of middle subarray
            for (int i = l + 1; i <= r - 1; i++) {
                msum += arr[i];
            }
 
            // Sum of right subarray
            for (int i = r + 1; i < N; i++) {
                rsum += arr[i];
            }
 
            // Check if sum of subarrays are equal
            if (lsum == rsum && rsum == msum) {
 
                // Print the possible pair
                System.out.println( l + " " + r );
                return;
            }
        }
    }
 
    // If no pair exists, print -1
    System.out.print(-1 );
}
 
// Driver Code
public static void main(String[] args)
{
   
    // Given array
    int arr[] = { 2, 5, 12, 7, 19, 4, 3 };
 
    // Size of the array
    int N = arr.length;
    findSplit(arr, N);
}
}
 
// This code is contributed by sanjoy_62.

Python3

# Pyhton 3 program for the above approach
 
# Function to check if array can
# be split into three equal sum
# subarrays by removing two elements
def findSplit(arr, N):
    for l in range(1, N - 3, 1):
        for r in range(l + 2, N - 1, 1):
           
            # Stores sum of all three subarrays
            lsum = 0
            rsum = 0
            msum = 0
 
            # Sum of left subarray
            for i in range(0, l, 1):
                lsum += arr[i]
 
            # Sum of middle subarray
            for i in range(l + 1, r, 1):
                msum += arr[i]
 
            # Sum of right subarray
            for i in range(r + 1, N, 1):
                rsum += arr[i]
 
            # Check if sum of subarrays are equal
            if (lsum == rsum and rsum == msum):
               
                # Print the possible pair
                print(l, r)
                return
 
    # If no pair exists, print -1
    print(-1)
 
# Driver code
if __name__ == '__main__':
   
    # Given array
    arr =  [2, 5, 12, 7, 19, 4, 3]
     
    # Size of the array
    N = len(arr)
    findSplit(arr, N)
     
    # This code is contributed by SURENDRA_GANGWAR.

C#

// C# program for the above approach
using System;
class GFG
{
 
  // Function to check if array can
  // be split into three equal sum
  // subarrays by removing two elements
  static void findSplit(int []arr, int N)
  {
    for (int l = 1; l <= N - 4; l++)
    {
      for (int r = l + 2; r <= N - 2; r++)
      {
 
        // Stores sum of all three subarrays
        int lsum = 0, rsum = 0, msum = 0;
 
        // Sum of left subarray
        for (int i = 0; i <= l - 1; i++) {
          lsum += arr[i];
        }
 
        // Sum of middle subarray
        for (int i = l + 1; i <= r - 1; i++) {
          msum += arr[i];
        }
 
        // Sum of right subarray
        for (int i = r + 1; i < N; i++) {
          rsum += arr[i];
        }
 
        // Check if sum of subarrays are equal
        if (lsum == rsum && rsum == msum) {
 
          // Print the possible pair
          Console.WriteLine( l + " " + r );
          return;
        }
      }
    }
 
    // If no pair exists, print -1
    Console.Write(-1 );
  }
 
  // Driver Code
  public static void Main(string[] args)
  {
 
    // Given array
    int []arr = { 2, 5, 12, 7, 19, 4, 3 };
 
    // Size of the array
    int N = arr.Length;
    findSplit(arr, N);
  }
}
 
// This code is contributed by AnkThon

输出：

2 4

时间复杂度： O(N ³ )
辅助空间： O(1)

高效的方法：为了优化上述方法，我们的想法是使用Prefix Sum数组技术在恒定时间内查找所有子数组和。请按照以下步骤解决问题：

• 初始化大小为N的向量和以存储数组的前缀和。
• 初始化两个变量，例如l ＆ r，以存储要删除的两个索引，以便将数组拆分为3个相等的子数组。
• 这三个子数组的总和为sum [l – 1] ， sum [r – 1] – sum [l]和sum [N – 1] – sum [r] 。
• 使用变量l在[1，N – 4]范围内迭代：
  • 使用变量r遍历[l + 2，N – 2]的范围，并检查在任何点上，左子数组之和等于中间子数组之和，右子数组之和，然后打印l和r的值并返回。
• 如果不存在这样的对，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 check if array can
// be split into three equal sum
// subarrays by removing a pair
void findSplit(int arr[], int N)
{
    // Stores prefix sum array
    vector sum(N);
 
    // Copy array elements
    for (int i = 0; i < N; i++) {
        sum[i] = arr[i];
    }
 
    // Traverse the array
    for (int i = 1; i < N; i++) {
        sum[i] += sum[i - 1];
    }
 
    for (int l = 1; l <= N - 4; l++) {
 
        for (int r = l + 2; r <= N - 2; r++) {
 
            // Stores sums of all three subarrays
            int lsum = 0, rsum = 0, msum = 0;
 
            // Sum of left subarray
            lsum = sum[l - 1];
 
            // Sum of middle subarray
            msum = sum[r - 1] - sum[l];
 
            // Sum of right subarray
            rsum = sum[N - 1] - sum[r];
 
            // Check if sum of subarrays are equal
            if (lsum == rsum && rsum == msum) {
 
                // Print the possible pair
                cout << l << " " << r << endl;
                return;
            }
        }
    }
 
    // If no such pair exists, print -1
    cout << -1 << endl;
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { 2, 5, 12, 7, 19, 4, 3 };
 
    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    findSplit(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG
{
 
// Function to check if array can
// be split into three equal sum
// subarrays by removing a pair
static void findSplit(int arr[], int N)
{
   
    // Stores prefix sum array
    int []sum = new int[N];
 
    // Copy array elements
    for (int i = 0; i < N; i++)
    {
        sum[i] = arr[i];
    }
 
    // Traverse the array
    for (int i = 1; i < N; i++)
    {
        sum[i] += sum[i - 1];
    }
 
    for (int l = 1; l <= N - 4; l++) {
 
        for (int r = l + 2; r <= N - 2; r++) {
 
            // Stores sums of all three subarrays
            int lsum = 0, rsum = 0, msum = 0;
 
            // Sum of left subarray
            lsum = sum[l - 1];
 
            // Sum of middle subarray
            msum = sum[r - 1] - sum[l];
 
            // Sum of right subarray
            rsum = sum[N - 1] - sum[r];
 
            // Check if sum of subarrays are equal
            if (lsum == rsum && rsum == msum) {
 
                // Print the possible pair
                System.out.print(l+ " " +  r +"\n");
                return;
            }
        }
    }
 
    // If no such pair exists, print -1
    System.out.print(-1 +"\n");
}
 
// Driver Code
public static void main(String[] args)
{
   
    // Given array
    int arr[] = { 2, 5, 12, 7, 19, 4, 3 };
 
    // Size of the array
    int N = arr.length;
    findSplit(arr, N);
}
}
 
// This code is contributed by shikhasingrajput

C＃

// C# program for the above approach
using System;
public class GFG
{
 
// Function to check if array can
// be split into three equal sum
// subarrays by removing a pair
static void findSplit(int []arr, int N)
{
   
    // Stores prefix sum array
    int []sum = new int[N];
 
    // Copy array elements
    for (int i = 0; i < N; i++)
    {
        sum[i] = arr[i];
    }
 
    // Traverse the array
    for (int i = 1; i < N; i++)
    {
        sum[i] += sum[i - 1];
    }
 
    for (int l = 1; l <= N - 4; l++) {
        for (int r = l + 2; r <= N - 2; r++) {
 
            // Stores sums of all three subarrays
            int lsum = 0, rsum = 0, msum = 0;
 
            // Sum of left subarray
            lsum = sum[l - 1];
 
            // Sum of middle subarray
            msum = sum[r - 1] - sum[l];
 
            // Sum of right subarray
            rsum = sum[N - 1] - sum[r];
 
            // Check if sum of subarrays are equal
            if (lsum == rsum && rsum == msum) {
 
                // Print the possible pair
                Console.Write(l+ " " +  r +"\n");
                return;
            }
        }
    }
 
    // If no such pair exists, print -1
    Console.Write(-1 +"\n");
}
 
// Driver Code
public static void Main(String[] args)
{
   
    // Given array
    int []arr = { 2, 5, 12, 7, 19, 4, 3 };
 
    // Size of the array
    int N = arr.Length;
    findSplit(arr, N);
}
}
 
// This code is contributed by 29AjayKumar

输出：

2 4

时间复杂度： O(N ² )
辅助空间： O(N)

最佳方法：最佳方法是同时使用两点技术和前缀和。请按照以下步骤解决问题：

• 初始化大小为N的向量以存储数组的前缀和。
• 初始化两个变量，例如l ＆ r ，以使用两指针方法遍历数组。
• 遍历数组直到l 或直到三个和都相等：
  • 如果左子数组的总和大于右子数组的总和，请向右子数组添加一个额外的元素。因此，将r的值减少1 。
  • 如果右子数组的总和大于左子数组的总和，则向左子数组添加一个元素。因此，将l增加1 。
  • 如果左右两个子数组的总和相等，但不等于中间子数组的总和，则将l增加1 ，将r减少1 。
• 如果不存在这样的对，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 check if array can
// be split into three equal sum
// subarrays by removing a pair
void findSplit(int arr[], int N)
{
    // Two pointers l and r
    int l = 1, r = N - 2;
    int lsum, msum, rsum;
 
    // Stores prefix sum array
    vector sum(N);
    sum[0] = arr[0];
 
    // Traverse the array
    for (int i = 1; i < N; i++) {
        sum[i] = sum[i - 1] + arr[i];
    }
 
    // Two pointer approach
    while (l < r) {
 
        // Sum of left subarray
        lsum = sum[l - 1];
 
        // Sum of middle subarray
        msum = sum[r - 1] - sum[l];
 
        // Sum of right subarray
        rsum = sum[N - 1] - sum[r];
 
        // Print split indices if sum is equal
        if (lsum == msum and msum == rsum) {
            cout << l << " " << r << endl;
            return;
        }
 
        // Move left pointer if lsum < rsum
        if (lsum < rsum)
            l++;
 
        // Move right pointer if rsum > lsum
        else if (lsum > rsum)
            r--;
 
        // Move both pointers if lsum = rsum
        // but they are not equal to msum
        else {
            l++;
            r--;
        }
    }
 
    // If no possible pair exists, print -1
    cout << -1 << endl;
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { 2, 5, 12, 7, 19, 4, 3 };
 
    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    findSplit(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
public class GFG
{
 
  // Function to check if array can
  // be split into three equal sum
  // subarrays by removing a pair
  static void findSplit(int []arr, int N)
  {
 
    // Two pointers l and r
    int l = 1, r = N - 2;
    int lsum, msum, rsum;
 
    // Stores prefix sum array
    int sum[] = new int[N];
 
    sum[0] = arr[0];
 
    // Traverse the array
    for (int i = 1; i < N; i++) {
      sum[i] = sum[i - 1] + arr[i];
    }
 
    // Two pointer approach
    while (l < r) {
 
      // Sum of left subarray
      lsum = sum[l - 1];
 
      // Sum of middle subarray
      msum = sum[r - 1] - sum[l];
 
      // Sum of right subarray
      rsum = sum[N - 1] - sum[r];
 
      // Print split indices if sum is equal
      if (lsum == msum && msum == rsum) {
        System.out.println(l + " " + r);
        return;
      }
 
      // Move left pointer if lsum < rsum
      if (lsum < rsum)
        l++;
 
      // Move right pointer if rsum > lsum
      else if (lsum > rsum)
        r--;
 
      // Move both pointers if lsum = rsum
      // but they are not equal to msum
      else {
        l++;
        r--;
      }
    }
 
    // If no possible pair exists, print -1
    System.out.println(-1);
  }
 
  // Driver Code
  public static void main (String[] args)
  {
    // Given array
    int []arr = { 2, 5, 12, 7, 19, 4, 3 };
 
    // Size of the array
    int N = arr.length;
 
    findSplit(arr, N);
  }
}
 
// This code is contributed by AnkThon

Python3

# Python3 program for the above approach
 
# Function to check if array can
# be split into three equal sum
# subarrays by removing a pair
def findSplit(arr, N) :
 
    # Two pointers l and r
    l = 1; r = N - 2;
 
    # Stores prefix sum array
    sum = [0]*N;
    sum[0] = arr[0];
 
    # Traverse the array
    for i in range(1, N) :
        sum[i] = sum[i - 1] + arr[i];
 
    # Two pointer approach
    while (l < r) :
 
        # Sum of left subarray
        lsum = sum[l - 1];
 
        # Sum of middle subarray
        msum = sum[r - 1] - sum[l];
 
        # Sum of right subarray
        rsum = sum[N - 1] - sum[r];
 
        # Print split indices if sum is equal
        if (lsum == msum and msum == rsum) :
            print(l,r);
            return;
 
        # Move left pointer if lsum < rsum
        if (lsum < rsum) :
            l += 1;
 
        # Move right pointer if rsum > lsum
        elif (lsum > rsum) :
            r -= 1;
 
        # Move both pointers if lsum = rsum
        # but they are not equal to msum
        else :
            l += 1;
            r -= 1;
 
    # If no possible pair exists, print -1
    print(-1);
 
# Driver Code
if __name__ == "__main__" :
 
    # Given array
    arr = [ 2, 5, 12, 7, 19, 4, 3 ];
 
    # Size of the array
    N = len(arr);
 
    findSplit(arr, N);
 
    # This code is contributed by AnkThon

输出：

2 4

时间复杂度： O(N)
辅助空间： O(N)