将所有元素替换为该元素后的正负数总和之差

给定一系列正负元素。任务是用i + 1到N范围内的正负元素的绝对和的绝对差值替换数组的第i个元素。即，在i + 1到N的范围内找到所有正元素的绝对和和所有负元素的绝对和。现在找到这两个和的绝对差，并用第i个元素替换。

注意：更新后的数组的最后一个元素将为零。

例子：

Input : N = 5,  arr[] = {1, -1, 2, 3, -2}
Output : arr[] = {2, 3, 1, 2, 0}

Input : N = 6,  arr[] = {-3, -4, -2, 5, 1, -2}
Output : arr[] = {2, 2, 4, 1, 2, 0}.

天真的方法：天真的方法是运行两个for循环，并对所有第i个元素，计算索引范围为i + 1到N的所有正负元素之和的abs值。现在找出两者的绝对差求和并用第i个元素替换。
这种方法的时间复杂度将是O(N ² )，其中N是数组中元素的数量。

下面是上述方法的实现：

C++

// C++ program to implement above approach
 
#include 
using namespace std;
 
// Function to print the array elements
void printArray(int N, int arr[])
{
    for (int i = 0; i < N; i++)
        cout << arr[i] << " ";
 
    cout << endl;
}
 
// Function to replace all elements with absolute
// difference of absolute sums of positive
// and negatve elements
void replacedArray(int N, int arr[])
{
    int pos_sum, neg_sum, i, j, diff;
 
    for (i = 0; i < N; i++) {
        pos_sum = 0;
        neg_sum = 0;
 
        // Calculate absolute sums of possitive
        // and negative elements in range i+1 to N
        for (j = i + 1; j < N; j++) {
            if (arr[j] > 0)
                pos_sum += arr[j];
            else
                neg_sum += arr[j];
        }
 
        // calculate difference of both sums
        diff = abs(pos_sum) - abs(neg_sum);
 
        // replace i-th elements with absolute
        // difference
        arr[i] = abs(diff);
    }
}
 
// Driver code
int main()
{
    int N = 5;
    int arr[] = { 1, -1, 2, 3, -2 };
    replacedArray(N, arr);
    printArray(N, arr);
 
    N = 6;
    int arr1[] = { -3, -4, -2, 5, 1, -2 };
    replacedArray(N, arr1);
    printArray(N, arr1);
 
    return 0;
}

Java

// Java program to implement above approach
class GFG
{
     
// Function to print the array elements
static void printArray(int N, int []arr)
{
    for (int i = 0; i < N; i++)
        System.out.print(arr[i] + " ");
 
    System.out.println();
}
 
// Function to replace all elements with
// absolute difference of absolute sums
// of positive and negatve elements
static void replacedArray(int N, int []arr)
{
    int pos_sum, neg_sum, i, j, diff;
 
    for (i = 0; i < N; i++)
    {
        pos_sum = 0;
        neg_sum = 0;
 
        // Calculate absolute sums of possitive
        // and negative elements in range i+1 to N
        for (j = i + 1; j < N; j++)
        {
            if (arr[j] > 0)
                pos_sum += arr[j];
            else
                neg_sum += arr[j];
        }
 
        // calculate difference of both sums
        diff = Math.abs(pos_sum) - Math.abs(neg_sum);
 
        // replace i-th elements with absolute
        // difference
        arr[i] = Math.abs(diff);
    }
}
 
// Driver code
public static void main(String args[])
{
    int N = 5;
    int []arr = { 1, -1, 2, 3, -2 };
    replacedArray(N, arr);
    printArray(N, arr);
 
    N = 6;
    int []arr1 = { -3, -4, -2, 5, 1, -2 };
    replacedArray(N, arr1);
    printArray(N, arr1);
}
}
 
// This code is contributed by Akanksha Rai

Python3

# Python 3 program to implement
# above approach
 
# Function to print the array elements
def printArray(N, arr):
    for i in range(N):
        print(arr[i], end = " ")
 
    print("\n", end = "")
 
# Function to replace all elements with
# absolute difference of absolute sums
# of positive and negatve elements
def replacedArray(N, arr):
    for i in range(N):
        pos_sum = 0
        neg_sum = 0
 
        # Calculate absolute sums of possitive
        # and negative elements in range i+1 to N
        for j in range(i + 1, N, 1):
            if (arr[j] > 0):
                pos_sum += arr[j]
            else:
                neg_sum += arr[j]
 
        # calculate difference of both sums
        diff = abs(pos_sum) - abs(neg_sum)
 
        # replace i-th elements with absolute
        # difference
        arr[i] = abs(diff)
 
# Driver code
if __name__ == '__main__':
    N = 5
    arr = [1, -1, 2, 3, -2]
    replacedArray(N, arr)
    printArray(N, arr)
 
    N = 6
    arr1 = [-3, -4, -2, 5, 1, -2]
    replacedArray(N, arr1)
    printArray(N, arr1)
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# program to implement above approach
using System;
 
class GFG
{
     
// Function to print the array elements
static void printArray(int N, int []arr)
{
    for (int i = 0; i < N; i++)
        Console.Write(arr[i] + " ");
 
    Console.WriteLine();
}
 
// Function to replace all elements with
// absolute difference of absolute sums
// of positive and negatve elements
static void replacedArray(int N, int []arr)
{
    int pos_sum, neg_sum, i, j, diff;
 
    for (i = 0; i < N; i++)
    {
        pos_sum = 0;
        neg_sum = 0;
 
        // Calculate absolute sums of possitive
        // and negative elements in range i+1 to N
        for (j = i + 1; j < N; j++)
        {
            if (arr[j] > 0)
                pos_sum += arr[j];
            else
                neg_sum += arr[j];
        }
 
        // calculate difference of both sums
        diff = Math.Abs(pos_sum) - Math.Abs(neg_sum);
 
        // replace i-th elements with absolute
        // difference
        arr[i] = Math.Abs(diff);
    }
}
 
// Driver code
static void Main()
{
    int N = 5;
    int []arr = { 1, -1, 2, 3, -2 };
    replacedArray(N, arr);
    printArray(N, arr);
 
    N = 6;
    int []arr1 = { -3, -4, -2, 5, 1, -2 };
    replacedArray(N, arr1);
    printArray(N, arr1);
}
}
 
// This code is contributed by mits

Javascript

 
// Javascript program to implement above approach   
 
// Function to prvar the array elements
function printArray(N, arr)
{
    for(i = 0; i < N; i++)
        document.write(arr[i] + " ");
 
    document.write("<br/>");
}
 
// Function to replace all elements with
// absolute difference of absolute sums
// of positive and negatve elements
function replacedArray(N, arr)
{
    var pos_sum, neg_sum, i, j, diff;
 
    for(i = 0; i < N; i++)
    {
        pos_sum = 0;
        neg_sum = 0;
 
        // Calculate absolute sums of possitive
        // and negative elements in range i+1 to N
        for(j = i + 1; j < N; j++)
        {
            if (arr[j] > 0)
                pos_sum += arr[j];
            else
                neg_sum += arr[j];
        }
 
        // Calculate difference of both sums
        diff = Math.abs(pos_sum) -
               Math.abs(neg_sum);
 
        // Replace i-th elements with absolute
        // difference
        arr[i] = Math.abs(diff);
    }
}
 
// Driver code
var N = 5;
var arr = [ 1, -1, 2, 3, -2 ];
replacedArray(N, arr);
printArray(N, arr);
 
N = 6;
var arr1 = [ -3, -4, -2, 5, 1, -2 ];
replacedArray(N, arr1);
printArray(N, arr1);
 
// This code is contributed by aashish1995
 

输出：

2 3 1 2 0 
2 2 4 1 2 0

高效方法：将正和负的和初始化为0。现在，从最后一个元素到第一个元素运行一个for循环，并计算diff = abs(pos_sum)– abs(neg_sum)。
现在，如果第i个元素为正，则将其添加到pos_sum，否则将其添加到neg_sum。毕竟，用绝对差即abs(diff)替换第i个元素。

下面是上述方法的实现：

C++

// C++ program to implement above approach
 
#include 
using namespace std;
 
// Function to print the array elements
void printArray(int N, int arr[])
{
    for (int i = 0; i < N; i++)
        cout << arr[i] << " ";
 
    cout << endl;
}
 
// Function to replace all elements with absolute
// difference of absolute sums of positive
// and negatve elements
void replacedArray(int N, int arr[])
{
    int pos_sum, neg_sum, i, j, diff;
 
    pos_sum = 0;
    neg_sum = 0;
 
    for (i = N - 1; i >= 0; i--) {
 
        // calculate differenbce of both sums
        diff = abs(pos_sum) - abs(neg_sum);
 
        // if i-th element is positive,
        // add it to positive sum
        if (arr[i] > 0)
            pos_sum += arr[i];
 
        // if i-th element is negative,
        // add it to negative sum
        else
            neg_sum += arr[i];
 
        // replace i-th elements with
        // absolute difference
        arr[i] = abs(diff);
    }
}
 
// Driver Code
int main()
{
    int N = 5;
    int arr[] = { 1, -1, 2, 3, -2 };
    replacedArray(N, arr);
    printArray(N, arr);
 
    N = 6;
    int arr1[] = { -3, -4, -2, 5, 1, -2 };
    replacedArray(N, arr1);
    printArray(N, arr1);
 
    return 0;
}

Java

// Java program to implement above approach
class GFG
{
     
    // Function to print the array elements
    static void printArray(int N, int arr[])
    {
        for (int i = 0; i < N; i++)
            System.out.print(arr[i] + " ");
     
        System.out.println();
    }
     
    // Function to replace all elements with absolute
    // difference of absolute sums of positive
    // and negatve elements
    static void replacedArray(int N, int arr[])
    {
        int pos_sum, neg_sum, i, j, diff;
     
        pos_sum = 0;
        neg_sum = 0;
     
        for (i = N - 1; i >= 0; i--)
        {
     
            // calculate differenbce of both sums
            diff = Math.abs(pos_sum) - Math.abs(neg_sum);
     
            // if i-th element is positive,
            // add it to positive sum
            if (arr[i] > 0)
                pos_sum += arr[i];
     
            // if i-th element is negative,
            // add it to negative sum
            else
                neg_sum += arr[i];
     
            // replace i-th elements with
            // absolute difference
            arr[i] = Math.abs(diff);
        }
    }
     
    // Driver Code
    public static void main (String[] args)
    {
        int N = 5;
        int arr[] = { 1, -1, 2, 3, -2 };
        replacedArray(N, arr);
        printArray(N, arr);
     
        N = 6;
        int arr1[] = { -3, -4, -2, 5, 1, -2 };
        replacedArray(N, arr1);
        printArray(N, arr1);
    }
}
 
// This code is contributed by ihritik

Python3

# Python program to implement above approach
 
# Function to print the array elements
def printArray(N, arr) :
 
    for i in range (0, N) :
        print(arr[i], end=" ")
 
    print()
 
 
# Function to replace all elements with absolute
# difference of absolute sums of positive
# and negatve elements
def replacedArray(N, arr) :
 
     
    pos_sum = 0
    neg_sum = 0
 
    for i in range (N - 1,-1, -1) :
 
        # calculate differenbce of both sums
        diff = abs(pos_sum) - abs(neg_sum)
 
        # if i-th element is positive,
        # add it to positive sum
        if (arr[i] > 0) :
            pos_sum = pos_sum + arr[i]
 
        # if i-th element is negative,
        # add it to negative sum
        else :
            neg_sum = neg_sum + arr[i]
 
        # replace i-th elements with
        # absolute difference
        arr[i] = abs(diff)
 
# Driver Code
 
N = 5
arr = [ 1, -1, 2, 3, -2 ]
replacedArray(N, arr)
printArray(N, arr)
 
N = 6
arr1 = [ -3, -4, -2, 5, 1, -2 ]
replacedArray(N, arr1)
printArray(N, arr1)
 
# This code is contributed by ihritik

C＃

// C# program to implement above approach
using System;
 
class GFG
{
     
    // Function to print the array elements
    static void printArray(int N, int [] arr)
    {
        for (int i = 0; i < N; i++)
            Console.Write(arr[i] + " ");
     
        Console.WriteLine();
    }
     
    // Function to replace all elements with absolute
    // difference of absolute sums of positive
    // and negatve elements
    static void replacedArray(int N, int [] arr)
    {
        int pos_sum, neg_sum, i, diff;
     
        pos_sum = 0;
        neg_sum = 0;
     
        for (i = N - 1; i >= 0; i--)
        {
     
            // calculate differenbce of both sums
            diff = Math.Abs(pos_sum) - Math.Abs(neg_sum);
     
            // if i-th element is positive,
            // add it to positive sum
            if (arr[i] > 0)
                pos_sum += arr[i];
     
            // if i-th element is negative,
            // add it to negative sum
            else
                neg_sum += arr[i];
     
            // replace i-th elements with
            // absolute difference
            arr[i] = Math.Abs(diff);
        }
    }
     
    // Driver Code
    public static void Main ()
    {
        int N = 5;
        int [] arr = { 1, -1, 2, 3, -2 };
        replacedArray(N, arr);
        printArray(N, arr);
     
        N = 6;
        int [] arr1 = { -3, -4, -2, 5, 1, -2 };
        replacedArray(N, arr1);
        printArray(N, arr1);
    }
}
 
// This code is contributed by ihritik

输出：

2 3 1 2 0 
2 2 4 1 2 0

时间复杂度： O(N)，其中N是元素数。