找到索引，其左侧的元素总数等于其右侧的元素总数之和

给定一个数组， arr []的大小为N ，任务是找到数组的最小索引，以使索引左侧的所有数组元素的总和等于所有n的总和的倒数。该索引右侧的数组元素。如果找不到这样的索引，则打印-1 。

例子：

Input: arr[] = { 5, 7, 3, 6, 4, 9, 2 } 
Output: 2 
Explanation: 
Sum of array elements on left side of index 2 = (5 + 7) = 12 
Sum of array elements on right side of index 2 = (6 + 4 + 9 + 2) = 21 
Since sum of array elements on left side of index 2 equal to the reverse of the digits of sum of elements on the right side of index 2. Therefore, the required output is 2.

Input: arr[] = { 1, 3, 6, 8, 9, 2 } 
Output: -1

天真的方法：解决此问题的最简单方法是遍历数组，对于每个索引，请检查索引左侧的元素总和是否等于数组左侧元素的总和的倒数。该索引的右侧与否。如果发现为真，则打印该索引。否则，如果找不到这样的索引，则打印-1 。

时间复杂度： O(N ² )
辅助空间： O(1)

高效的方法：为了优化上述方法，其思想是使用前缀和技术。请按照以下步骤解决问题：

• 初始化一个变量rightSum ，以将数组元素的总和存储在每个索引的右侧。
• 初始化一个变量，例如leftSum ，以将数组元素的总和存储在每个索引的左侧。
• 使用变量i遍历数组，并更新rightSum-= arr [i]，leftSum + = arr [i]的值，并检查leftSum是否等于rightSum的数字的倒数。如果发现是真的，则打印i 。
• 否则，打印-1 。

下面是上述方法的实现。

C++

// C++ Program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to check if a number is equal
// to the reverse of digits of other number
bool checkReverse(int leftSum, int rightSum)
{
    // Stores reverse of
    // digits of rightSum
    int rev = 0;
 
    // Stores rightSum
    int temp = rightSum;
 
    // Calculate reverse of
    // digits of temp
    while (temp != 0) {
 
        // Update rev
        rev = (rev * 10) + (temp % 10);
 
        // Update temp
        temp /= 10;
    }
 
    // If reverse of digits of
    // rightSum equal to leftSum
    if (rev == leftSum) {
        return true;
    }
 
    return false;
}
 
// Function to find the index
// that satisfies the condition
int findIndex(int arr[], int N)
{
 
    // Stores sum of array elements
    // on right side of each index
    int rightSum = 0;
 
    // Stores sum of array elements
    // on left side of each index
    int leftSum = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Update rightSum
        rightSum += arr[i];
    }
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Update rightSum
        rightSum -= arr[i];
 
        // If leftSum equal to
        // reverse of digits
        // of rightSum
        if (checkReverse(leftSum,
                         rightSum)) {
            return i;
        }
 
        // Update leftSum
        leftSum += arr[i];
    }
 
    return -1;
}
 
// Driver Code
int main()
{
    int arr[] = { 5, 7, 3, 6, 4, 9, 2 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << findIndex(arr, N);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG {
 
  // Function to check if a number is equal
  // to the reverse of digits of other number
  static boolean checkReverse(int leftSum, int rightSum)
  {
    // Stores reverse of
    // digits of rightSum
    int rev = 0;
 
    // Stores rightSum
    int temp = rightSum;
 
    // Calculate reverse of
    // digits of temp
    while (temp != 0) {
 
      // Update rev
      rev = (rev * 10) + (temp % 10);
 
      // Update temp
      temp /= 10;
    }
 
    // If reverse of digits of
    // rightSum equal to leftSum
    if (rev == leftSum) {
      return true;
    }
 
    return false;
  }
 
  // Function to find the index
  // that satisfies the condition
  static int findIndex(int[] arr, int N)
  {
 
    // Stores sum of array elements
    // on right side of each index
    int rightSum = 0;
 
    // Stores sum of array elements
    // on left side of each index
    int leftSum = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
      // Update rightSum
      rightSum += arr[i];
    }
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
      // Update rightSum
      rightSum -= arr[i];
 
      // If leftSum equal to
      // reverse of digits
      // of rightSum
      if (checkReverse(leftSum,
                       rightSum)) {
        return i;
      }
 
      // Update leftSum
      leftSum += arr[i];
    }
 
    return -1;
  }
 
  // Driver code
  public static void main(String[] args)
  {
 
    int[] arr = { 5, 7, 3, 6, 4, 9, 2 };
    int N = arr.length;
    System.out.print(findIndex(arr, N));
  }
}
 
// This code is contributed by code_hunt.

Python3

# Python3 Program to implement
# the above approach
 
# Function to check if a number is equal
# to the reverse of digits of other number
def checkReverse(leftSum, rightSum):
     
    # Stores reverse of
    # digits of rightSum
    rev = 0
 
    # Stores rightSum
    temp = rightSum
 
    # Calculate reverse of
    # digits of temp
    while (temp != 0):
 
        # Update rev
        rev = (rev * 10) + (temp % 10)
 
        # Update temp
        temp //= 10
 
    # If reverse of digits of
    # rightSum equal to leftSum
    if (rev == leftSum):
        return True
 
    return False
 
# Function to find the index
# that satisfies the condition
def findIndex(arr, N):
 
    # Stores sum of array elements
    # on right side of each index
    rightSum = 0
 
    # Stores sum of array elements
    # on left side of each index
    leftSum = 0
 
    # Traverse the array
    for i in range(N):
        
        # Update rightSum
        rightSum += arr[i]
 
    # Traverse the array
    for i in range(N):
 
        # Update rightSum
        rightSum -= arr[i]
 
        # If leftSum equal to
        # reverse of digits
        # of rightSum
        if (checkReverse(leftSum, rightSum)):
            return i
 
        # Update leftSum
        leftSum += arr[i]
    return -1
 
# Driver Code
if __name__ == '__main__':
    arr = [5, 7, 3, 6, 4, 9, 2]
    N =  len(arr)
    print(findIndex(arr, N))
 
    # This code is contributed by mohit kumar 29

C#

// C# Program to implement
// the above approach
using System;
class GFG {
     
    // Function to check if a number is equal
    // to the reverse of digits of other number
    static bool checkReverse(int leftSum, int rightSum)
    {
        // Stores reverse of
        // digits of rightSum
        int rev = 0;
      
        // Stores rightSum
        int temp = rightSum;
      
        // Calculate reverse of
        // digits of temp
        while (temp != 0) {
      
            // Update rev
            rev = (rev * 10) + (temp % 10);
      
            // Update temp
            temp /= 10;
        }
      
        // If reverse of digits of
        // rightSum equal to leftSum
        if (rev == leftSum) {
            return true;
        }
      
        return false;
    }
      
    // Function to find the index
    // that satisfies the condition
    static int findIndex(int[] arr, int N)
    {
      
        // Stores sum of array elements
        // on right side of each index
        int rightSum = 0;
      
        // Stores sum of array elements
        // on left side of each index
        int leftSum = 0;
      
        // Traverse the array
        for (int i = 0; i < N; i++) {
      
            // Update rightSum
            rightSum += arr[i];
        }
      
        // Traverse the array
        for (int i = 0; i < N; i++) {
      
            // Update rightSum
            rightSum -= arr[i];
      
            // If leftSum equal to
            // reverse of digits
            // of rightSum
            if (checkReverse(leftSum,
                             rightSum)) {
                return i;
            }
      
            // Update leftSum
            leftSum += arr[i];
        }
      
        return -1;
    }
 
  // Driver code
  static void Main()
  {
    int[] arr = { 5, 7, 3, 6, 4, 9, 2 };
    int N = arr.Length;
    Console.Write(findIndex(arr, N));
  }
}
 
// This code is contributed by divyeshrabadiya07

Javascript

 
// Javascript program for the above approach
 
    // Function to check if a number is equal
    // to the reverse of digits of other number
    function checkReverse(leftSum , rightSum) {
        // Stores reverse of
        // digits of rightSum
        var rev = 0;
 
        // Stores rightSum
        var temp = rightSum;
 
        // Calculate reverse of
        // digits of temp
        while (temp != 0) {
 
            // Update rev
            rev = (rev * 10) + (temp % 10);
 
            // Update temp
            temp = parseInt(temp/10);
        }
 
        // If reverse of digits of
        // rightSum equal to leftSum
        if (rev == leftSum) {
            return true;
        }
 
        return false;
    }
 
    // Function to find the index
    // that satisfies the condition
    function findIndex(arr , N) {
 
        // Stores sum of array elements
        // on right side of each index
        var rightSum = 0;
 
        // Stores sum of array elements
        // on left side of each index
        var leftSum = 0;
 
        // Traverse the array
        for (i = 0; i < N; i++) {
 
            // Update rightSum
            rightSum += arr[i];
        }
 
        // Traverse the array
        for (i = 0; i < N; i++) {
 
            // Update rightSum
            rightSum -= arr[i];
 
            // If leftSum equal to
            // reverse of digits
            // of rightSum
            if (checkReverse(leftSum, rightSum)) {
                return i;
            }
 
            // Update leftSum
            leftSum += arr[i];
        }
 
        return -1;
    }
 
    // Driver code
     
 
        var arr = [ 5, 7, 3, 6, 4, 9, 2 ];
        var N = arr.length;
        document.write(findIndex(arr, N));
 
// This code contributed by aashish1995
 

输出：

2

时间复杂度： O(N)
辅助空间： O(1)