重新排列数组以使从第一个索引非零开始的所有子数组的总和

给定一个由N 个整数组成的数组arr[] ，任务是重新排列数组，使得从数组的第一个索引开始的所有子数组的总和不为零。如果无法生成这样的排列，则打印“-1” 。

例子：

Input: arr[] = {-1, 1, -2, 3}
Output: {-1, -2, 1, 3}
Explanation: One of the possible rearrangement is {-1, -2, 1, 3}.
Subarrays starting from index 0 are {-1}, {-1, -2}, {-1, -2, 1} and {-1, -2, 1, 3}. None of the above subarrays have sum 0.

Input: arr[] = {0, 0, 0, 0}
Output: -1

方法：如果给定数组是以下两种配置中的任何一种，则可以从给定数组中获取所需数组：

• 如果给定数组按升序排序，则可以通过用大于它的元素替换子数组的最后一个元素来处理总和为零的第一个子数组。
• 类似地，在按降序排序的数组中，通过用比它小的元素替换和为零的子数组的第一个元素来确保其后的和为负。

请按照以下步骤解决问题：

1. 当数组按升序排序时：
   • 按升序对数组arr[]进行排序，并找到数组的前 i 个元素的总和 (0 ≤ i ≤ N)。
   • 当遇到零和时，用数组的最大元素替换使前缀和无效的元素（即^第i^个元素）：
     • 如果数组的最大元素等于导致无效的整数，则移动到第二个配置。
     • 如果最大元素大于有问题的元素，则此替换确保正和而不是零。
2. 当数组按降序排序时：
   • 按降序对数组arr[]进行排序并开始查找数组的最后 i 个元素的总和 (0 ≤ i ≤ N)。
   • 当遇到零和时，用数组的最小元素替换使前缀和无效的元素（即^第i^个元素）：
     • 如果数组的最小元素等于导致无效的整数，则无法重新排列数组 arr[]。
     • 如果最小元素小于有问题的元素，则此替换确保负和而不是零。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to rearrange the array such
// that sum of all elements of subarrays
// from the 1st index is non-zero
void rearrangeArray(int a[], int N)
{
    // Initialize sum of subarrays
    int sum = 0;
 
    // Sum of all elements of array
    for (int i = 0; i < N; i++) {
 
        sum += a[i];
    }
 
    // If sum is 0, the required
    // array could never be formed
    if (sum == 0) {
        cout << "-1";
        return;
    }
 
    // If sum is non zero, array
    // might be formed
    sum = 0;
 
    int b = 0;
 
    // Sort array in ascending order
    sort(a, a + N);
 
    for (int i = 0; i < N; i++) {
        sum += a[i];
 
        // When current subarray sum
        // becomes 0 replace it with
        // the largest element
        if (sum == 0) {
 
            if (a[i] != a[N - 1]) {
 
                sum -= a[i];
 
                // Swap Operation
                swap(a[i], a[N - 1]);
                sum += a[i];
            }
 
            // If largest element is same
            // as element to be replaced,
            // then rearrangement impossible
            else {
                b = 1;
                break;
            }
        }
    }
 
    // If b = 1, then rearrangement
    // is not possible. Hence check
    // with reverse configuration
    if (b == 1) {
 
        b = 0;
        sum = 0;
 
        // Sort array in descending order
        sort(a, a + N, greater());
 
        // When current subarray sum
        // becomes 0 replace it with
        // the smallest element
        for (int i = N - 1; i >= 0; i--) {
 
            sum += a[i];
            if (sum == 0) {
                if (a[i] != a[0]) {
                    sum -= a[i];
 
                    // Swap Operation
                    swap(a[i], a[0]);
                    sum += a[i];
                }
 
                // If smallest element is same
                // as element to be replaced,
                // then rearrangement impossible
                else {
                    b = 1;
                    break;
                }
            }
        }
    }
 
    // If neither of the configurations
    // worked then print "-1"
    if (b == 1) {
        cout << "-1";
        return;
    }
 
    // Otherwise, print the formed
    // rearrangement
    for (int i = 0; i < N; i++) {
        cout << a[i] << " ";
    }
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { 1, -1, 2, 4, 0 };
 
    // Size of array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    rearrangeArray(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
import java.util.Arrays;
import java.util.Collections;
 
class GFG{
 
// Function to rearrange the array such
// that sum of all elements of subarrays
// from the 1st index is non-zero
static void rearrangeArray(int a[], int N)
{
     
    // Initialize sum of subarrays
    int sum = 0;
 
    // Sum of all elements of array
    for(int i = 0; i < N; i++)
    {
        sum += a[i];
    }
 
    // If sum is 0, the required
    // array could never be formed
    if (sum == 0)
    {
        System.out.print("-1");
        return;
    }
 
    // If sum is non zero, array
    // might be formed
    sum = 0;
 
    int b = 0;
 
    // Sort array in ascending order
    Arrays.sort(a);
     
    for(int i = 0; i < N; i++)
    {
        sum += a[i];
 
        // When current subarray sum
        // becomes 0 replace it with
        // the largest element
        if (sum == 0)
        {
            if (a[i] != a[N - 1])
            {
                sum -= a[i];
                 
                // Swap Operation
                int temp = a[i];
                a[i] = a[N - 1];
                a[N - 1] = temp;
                sum += a[i];
            }
 
            // If largest element is same
            // as element to be replaced,
            // then rearrangement impossible
            else
            {
                b = 1;
                break;
            }
        }
    }
 
    // If b = 1, then rearrangement
    // is not possible. Hence check
    // with reverse configuration
    if (b == 1)
    {
        b = 0;
        sum = 0;
 
        // Sort array in descending order
        Arrays.sort(a);
 
        // When current subarray sum
        // becomes 0 replace it with
        // the smallest element
        for(int i = N - 1; i >= 0; i--)
        {
            sum += a[i];
            if (sum == 0)
            {
                if (a[i] != a[0])
                {
                    sum -= a[i];
                     
                    // Swap Operation
                    int temp = a[i];
                    a[i] = a[0];
                    a[0] = temp;
                    sum += a[i];
                }
 
                // If smallest element is same
                // as element to be replaced,
                // then rearrangement impossible
                else
                {
                    b = 1;
                    break;
                }
            }
        }
    }
 
    // If neither of the configurations
    // worked then print "-1"
    if (b == 1)
    {
        System.out.print("-1" + " ");
        return;
    }
 
    // Otherwise, print the formed
    // rearrangement
    for(int i = 0; i < N; i++)
    {
        System.out.print(a[i] + " ");
    }
}
 
// Driver Code
public static void main(String args[])
{
     
    // Given array
    int arr[] = { 1, -1, 2, 4, 0 };
 
    // Size of array
    int N = arr.length;
 
    // Function Call
    rearrangeArray(arr, N);
}
}
 
// This code is contributed by SURENDRA_GANGWAR

Python3

# Python3 program for the above approach
 
# Function to rearrange the array such
# that sum of all elements of subarrays
# from the 1st index is non-zero
def rearrangeArray(a, N):
     
    # Initialize sum of subarrays
    sum = 0
 
    # Sum of all elements of array
    for i in range(N):
        sum += a[i]
 
    # If sum is 0, the required
    # array could never be formed
    if (sum == 0):
        print("-1")
        return
 
    # If sum is non zero, array
    # might be formed
    sum = 0
 
    b = 0
     
    # Sort array in ascending order
    a = sorted(a)
 
    for i in range(N):
        sum += a[i]
 
        # When current subarray sum
        # becomes 0 replace it with
        # the largest element
        if (sum == 0):
            if (a[i] != a[N - 1]):
                sum -= a[i]
 
                # Swap Operation
                a[i], a[N - 1] = a[N - 1], a[i]
                sum += a[i]
 
            # If largest element is same
            # as element to be replaced,
            # then rearrangement impossible
            else:
                b = 1
                break
 
    # If b = 1, then rearrangement
    # is not possible. Hence check
    # with reverse configuration
    if (b == 1):
        b = 0
        sum = 0
 
        # Sort array in descending order
        a = sorted(a)
        a = a[::-1]
 
        # When current subarray sum
        # becomes 0 replace it with
        # the smallest element
        for i in range(N - 1, -1, -1):
            sum += a[i]
             
            if (sum == 0):
                if (a[i] != a[0]):
                    sum -= a[i]
 
                    # Swap Operation
                    a[i], a[0] = a[0], a[i]
                    sum += a[i]
 
                # If smallest element is same
                # as element to be replaced,
                # then rearrangement impossible
                else:
                    b = 1
                    break
 
    # If neither of the configurations
    # worked then pr"-1"
    if (b == 1):
        print("-1")
        return
 
    # Otherwise, print the formed
    # rearrangement
    for i in range(N):
        print(a[i], end = " ")
 
# Driver Code
if __name__ == '__main__':
     
    # Given array
    arr = [ 1, -1, 2, 4, 0 ]
 
    # Size of array
    N = len(arr)
 
    # Function Call
    rearrangeArray(arr, N)
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to rearrange the array such
// that sum of all elements of subarrays
// from the 1st index is non-zero
static void rearrangeArray(int [] a, int N)
{
     
    // Initialize sum of subarrays
    int sum = 0;
 
    // Sum of all elements of array
    for(int i = 0; i < N; i++)
    {
        sum += a[i];
    }
 
    // If sum is 0, the required
    // array could never be formed
    if (sum == 0)
    {
        Console.Write("-1");
        return;
    }
 
    // If sum is non zero, array
    // might be formed
    sum = 0;
 
    int b = 0;
 
    // Sort array in ascending order
    Array.Sort(a);
     
    for(int i = 0; i < N; i++)
    {
        sum += a[i];
         
        // When current subarray sum
        // becomes 0 replace it with
        // the largest element
        if (sum == 0)
        {
            if (a[i] != a[N - 1])
            {
                sum -= a[i];
                 
                // Swap Operation
                int temp = a[i];
                a[i] = a[N - 1];
                a[N - 1] = temp;
                sum += a[i];
            }
             
            // If largest element is same
            // as element to be replaced,
            // then rearrangement impossible
            else
            {
                b = 1;
                break;
            }
        }
    }
     
    // If b = 1, then rearrangement
    // is not possible. Hence check
    // with reverse configuration
    if (b == 1)
    {
        b = 0;
        sum = 0;
         
        // Sort array in descending order
        Array.Sort(a);
 
        // When current subarray sum
        // becomes 0 replace it with
        // the smallest element
        for(int i = N - 1; i >= 0; i--)
        {
            sum += a[i];
            if (sum == 0)
            {
                if (a[i] != a[0])
                {
                    sum -= a[i];
                     
                    // Swap Operation
                    int temp = a[i];
                    a[i] = a[0];
                    a[0] = temp;
                    sum += a[i];
                }
                 
                // If smallest element is same
                // as element to be replaced,
                // then rearrangement impossible
                else
                {
                    b = 1;
                    break;
                }
            }
        }
    }
     
    // If neither of the configurations
    // worked then print "-1"
    if (b == 1)
    {
        Console.Write("-1" + " ");
        return;
    }
     
    // Otherwise, print the formed
    // rearrangement
    for(int i = 0; i < N; i++)
    {
        Console.Write(a[i] + " ");
    }
}
 
// Driver Code
public static void Main()
{
     
    // Given array
    int[] arr = { 1, -1, 2, 4, 0 };
 
    // Size of array
    int N = arr.Length;
 
    // Function Call
    rearrangeArray(arr, N);
}
}
 
// This code is contributed by chitranayal

Javascript

// javascript program for the
// above approach
 
// Function to rearrange the array such
// that sum of all elements of subarrays
// from the 1st index is non-zero
function rearrangeArray(a, N)
{
      
    // Initialize sum of subarrays
    let sum = 0;
  
    // Sum of all elements of array
    for(let i = 0; i < N; i++)
    {
        sum += a[i];
    }
  
    // If sum is 0, the required
    // array could never be formed
    if (sum == 0)
    {
        document.write("-1");
        return;
    }
  
    // If sum is non zero, array
    // might be formed
    sum = 0;
  
    let b = 0;
  
    // Sort array in ascending order
    a.sort();
      
    for(let i = 0; i < N; i++)
    {
        sum += a[i];
  
        // When current subarray sum
        // becomes 0 replace it with
        // the largest element
        if (sum == 0)
        {
            if (a[i] != a[N - 1])
            {
                sum -= a[i];
                  
                // Swap Operation
                let temp = a[i];
                a[i] = a[N - 1];
                a[N - 1] = temp;
                sum += a[i];
            }
  
            // If largest element is same
            // as element to be replaced,
            // then rearrangement impossible
            else
            {
                b = 1;
                break;
            }
        }
    }
  
    // If b = 1, then rearrangement
    // is not possible. Hence check
    // with reverse configuration
    if (b == 1)
    {
        b = 0;
        sum = 0;
  
        // Sort array in descending order
        a.sort();
  
        // When current subarray sum
        // becomes 0 replace it with
        // the smallest element
        for(let i = N - 1; i >= 0; i--)
        {
            sum += a[i];
            if (sum == 0)
            {
                if (a[i] != a[0])
                {
                    sum -= a[i];
                      
                    // Swap Operation
                    let temp = a[i];
                    a[i] = a[0];
                    a[0] = temp;
                    sum += a[i];
                }
  
                // If smallest element is same
                // as element to be replaced,
                // then rearrangement impossible
                else
                {
                    b = 1;
                    break;
                }
            }
        }
    }
  
    // If neither of the configurations
    // worked then print "-1"
    if (b == 1)
    {
        document.write("-1" + " ");
        return;
    }
  
    // Otherwise, print the formed
    // rearrangement
    for(let i = 0; i < N; i++)
    {
        document.write(a[i] + " ");
    }
}
  
// Driver Code
 
    // Given array
    let arr = [ 1, -1, 2, 4, 0 ];
  
    // Size of array
    let N = arr.length;
  
    // Function Call
    rearrangeArray(arr, N);
           

输出：

-1 0 4 2 1

时间复杂度： O(N*log N)
辅助空间： O(1)

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