Find the Initial Array from given array after range sum queries

Last Updated : 31 Mar, 2023

Given an array arr[] which is the resultant array when a number of queries are performed on the original array. The queries are of the form [l, r, x] where l is the starting index in the array, r is the ending index in the array and x is the integer elements that have to be added to all the elements in the index range [l, r]. The task is to find the original array.

Examples:

Input: arr[] = {5, 7, 8}, l[] = {0}, r[] = {1}, x[] = {2}
Output: 3 5 8
If query [0, 1, 2] is performed on the array {3, 5, 8}
The resultant array will be {5, 7, 8}

Input: arr[] = {20, 30, 20, 70, 100},
l[] = {0, 1, 3},
r[] = {2, 4, 4},
x[] = {10, 20, 30}
Output: 10 0 -10 20 50

Naive Approach: For each range starting from l to r subtract the corresponding x to get the initial array.

Algorithm

1)Define the initial array arr and its size n.
2)Define the ranges l, r, and the values x for decrementing the elements in those ranges.
3)Define the number of queries q.
4)For each query j from 0 to q-1:
    For each index i from l[j] to r[j], decrement the corresponding element in the array arr by the value x[j].
5)Print the elements of the final array arr.

Below is the implementation of the approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Utility function to print the contents of an array
void printArr(int arr[], int n)
{
    for (int i = 0; i < n; i++) {
        cout << arr[i] << " ";
    }
}

// Function to find the original array
void findOrgArr(int arr[], int l[], int r[], int x[],
                int n, int q)
{
    for (int j = 0; j < q; j++) {
        for (int i = l[j]; i <= r[j]; i++) {

            // Decrement elements between
            // l[j] and r[j] by x[j]
            arr[i] = arr[i] - x[j];
        }
    }

    printArr(arr, n);
}

// Driver code
int main()
{
    // Final array
    int arr[] = { 20, 30, 20, 70, 100 };

    // Size of the array
    int n = sizeof(arr) / sizeof(arr[0]);

    // Queries
    int l[] = { 0, 1, 3 };
    int r[] = { 2, 4, 4 };
    int x[] = { 10, 20, 30 };

    // Number of queries
    int q = sizeof(l) / sizeof(l[0]);

    findOrgArr(arr, l, r, x, n, q);

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG
{

// Utility function to print the contents of an array
static void printArr(int arr[], int n)
{
    for (int i = 0; i < n; i++) {
        System.out.print(arr[i]+" ");
    }
}

// Function to find the original array
static void findOrgArr(int arr[], int l[], int r[], int x[],
                int n, int q)
{
    for (int j = 0; j < q; j++) {
        for (int i = l[j]; i <= r[j]; i++) {

            // Decrement elements between
            // l[j] and r[j] by x[j]
            arr[i] = arr[i] - x[j];
        }
    }

    printArr(arr, n);
}

// Driver code
public static void  main(String args[])
{
    // Final array
    int arr[] = { 20, 30, 20, 70, 100 };

    // Size of the array
    int n =  arr.length;

    // Queries
    int l[] = { 0, 1, 3 };
    int r[] = { 2, 4, 4 };
    int x[] = { 10, 20, 30 };

    // Number of queries
    int q = l.length;

    findOrgArr(arr, l, r, x, n, q);

}
}

// This code is contributed by
// Shashank_Sharma

Python3

# Python3 implementation of the approach
import math as mt

# Utility function to print the 
# contents of an array
def printArr(arr, n):

    for i in range(n): 
        print(arr[i], end = " ")

# Function to find the original array
def findOrgArr(arr, l, r, x, n, q):

    for j in range(q):
        for i in range(l[j], r[j] + 1):
            
            # Decrement elements between
            # l[j] and r[j] by x[j]
            arr[i] = arr[i] - x[j]
        
    printArr(arr, n)

# Driver code

# Final array
arr = [20, 30, 20, 70, 100] 

# Size of the array
n = len(arr)

# Queries
l = [0, 1, 3] 
r = [ 2, 4, 4] 
x = [ 10, 20, 30 ]

# Number of queries
q = len(l)

findOrgArr(arr, l, r, x, n, q)

# This code is contributed by 
# mohit kumar 29

// C# implementation of the approach
using System;

class GFG
{

// Utility function to print the 
// contents of an array
static void printArr(int[] arr, int n)
{
    for (int i = 0; i < n; i++)
    {
        Console.Write(arr[i] + " ");
    }
}

// Function to find the original array
static void findOrgArr(int[] arr, int[] l, 
                       int[] r, int[] x, 
                       int n, int q)
{
    for (int j = 0; j < q; j++) 
    {
        for (int i = l[j]; i <= r[j]; i++)
        {

            // Decrement elements between
            // l[j] and r[j] by x[j]
            arr[i] = arr[i] - x[j];
        }
    }

    printArr(arr, n);
}

// Driver code
public static void Main()
{
    // Final array
    int[] arr = { 20, 30, 20, 70, 100 };

    // Size of the array
    int n = arr.Length;

    // Queries
    int[] l = { 0, 1, 3 };
    int[] r = { 2, 4, 4 };
    int[] x = { 10, 20, 30 };

    // Number of queries
    int q = l.Length;

    findOrgArr(arr, l, r, x, n, q);

}
}

// This code is contributed by
// Akanksha Rai

PHP

<?php
// PHP implementation of the approach

// Utility function to print the contents
// of an array
function printArr(&$arr, $n)
{
    for ($i = 0; $i < $n; $i++)
    {
        echo($arr[$i]);
        echo(" ");
    }
}

// Function to find the original array
function findOrgArr(&$arr, &$l, &$r, 
                        &$x, $n, $q)
{
    for ($j = 0; $j < $q; $j++) 
    {
        for ($i = $l[$j]; $i <= $r[$j]; $i++) 
        {

            // Decrement elements between
            // l[j] and r[j] by x[j]
            $arr[$i] = $arr[$i] - $x[$j];
        }
    }

    printArr($arr, $n);
}

// Driver code

// Final array
$arr = array(20, 30, 20, 70, 100);

// Size of the array
$n = sizeof($arr);

// Queries
$l = array(0, 1, 3 );
$r = array( 2, 4, 4 );
$x = array(10, 20, 30 );

// Number of queries
$q = sizeof($l);

findOrgArr($arr, $l, $r, $x, $n, $q);

// This code is contributed by Shivi_Aggarwal
?>

JavaScript

<script>
// Javascript implementation of the approach

// Utility function to print the contents of an array
function printArr(arr,n)
{
    for (let i = 0; i < n; i++) {
        document.write(arr[i]+" ");
    }
}

// Function to find the original array
function findOrgArr(arr,l,r,x,n,q)
{
     for (let j = 0; j < q; j++) {
        for (let i = l[j]; i <= r[j]; i++) {
  
            // Decrement elements between
            // l[j] and r[j] by x[j]
            arr[i] = arr[i] - x[j];
        }
    }
  
    printArr(arr, n);
}

// Driver code

// Final array
let arr = [ 20, 30, 20, 70, 100 ];

// Size of the array
let n =  arr.length;

// Queries
let l = [ 0, 1, 3 ];
let r = [ 2, 4, 4 ];
let x = [ 10, 20, 30 ];

// Number of queries
let q = l.length;

findOrgArr(arr, l, r, x, n, q);

        
// This code is contributed by patel2127
</script>

Output

10 0 -10 20 50

Complexity Analysis:

Time Complexity: O(n²)
Auxiliary Space: O(1)

Efficient Approach:

Follow the following steps to reach the initial array:

Take an array b[] of the size of the given array and initialize all of its elements with 0.
In array b[], for every query update b[l] = b[l] - x and b[r + 1] = b[r + 1] + x if r + 1 < n. This is because x will cancel out the effect of -x when performed the prefix sum.
Take the prefix sum of array b[], and add it to the given array which will produce the initial array.

Implementation:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Utility function to print the contents of an array
void printArr(int arr[], int n)
{
    for (int i = 0; i < n; i++) {
        cout << arr[i] << " ";
    }
}

// Function to find the original array
void findOrgArr(int arr[], int l[], int r[], int x[],
                int n, int q)
{
    int b[n] = { 0 };

    for (int i = 0; i < q; i++) {

        // Decrement the element at l[i]th index by -x
        b[l[i]] += -x[i];

        // Increment the element at (r[i] + 1)th index
        // by x if (r[i] + 1) is a valid index
        if (r[i] + 1 < n)
            b[r[i] + 1] += x[i];
    }

    for (int i = 1; i < n; i++)
        // Prefix sum of array b
        b[i] = b[i - 1] + b[i];

    // Update the original array
    for (int i = 0; i < n; i++)
        arr[i] = arr[i] + b[i];

    printArr(arr, n);
}

// Driver code
int main()
{
    // Final array
    int arr[] = { 20, 30, 20, 70, 100 };

    // Size of the array
    int n = sizeof(arr) / sizeof(arr[0]);

    // Queries
    int l[] = { 0, 1, 3 };
    int r[] = { 2, 4, 4 };
    int x[] = { 10, 20, 30 };

    // Number of queries
    int q = sizeof(l) / sizeof(l[0]);

    findOrgArr(arr, l, r, x, n, q);

    return 0;
}

Java

// Java implementation of above approach 
class GFG{

    // Utility function to print the contents of an array 
    static void printArr(int arr[], int n) 
    { 
        for (int i = 0; i < n; i++) 
        { 
        System.out.print(arr[i] + " ") ; 
        } 
    } 
    
    // Function to find the original array 
    static void findOrgArr(int arr[], int l[], int r[], int x[], 
                    int n, int q) 
    { 
        int b[] = new int[n] ;
        
        for (int i = 0; i < q; i++)
            b[i] = 0 ;
    
        for (int i = 0; i < q; i++)
        { 
    
            // Decrement the element at l[i]th index by -x 
            b[l[i]] += -x[i]; 
    
            // Increment the element at (r[i] + 1)th index 
            // by x if (r[i] + 1) is a valid index 
            if (r[i] + 1 < n) 
                b[r[i] + 1] += x[i]; 
        } 
    
        for (int i = 1; i < n; i++) 
            // Prefix sum of array b 
            b[i] = b[i - 1] + b[i]; 
    
        // Update the original array 
        for (int i = 0; i < n; i++) 
            arr[i] = arr[i] + b[i]; 
    
        printArr(arr, n); 
    } 
    
    // Driver code 
    public static void main(String []args)
    { 
        // Final array 
        int arr[] = { 20, 30, 20, 70, 100 }; 
    
        // Size of the array 
        int n = arr.length ;
    
        // Queries 
        int l[] = { 0, 1, 3 }; 
        int r[] = { 2, 4, 4 }; 
        int x[] = { 10, 20, 30 }; 
    
        // Number of queries 
        int q = l.length ;
    
        findOrgArr(arr, l, r, x, n, q); 
        } 
}

// This code is contributed by aishwarya.27

Python3

# Python3 implementation of the approach

# Utility function to print the contents 
# of an array
def printArr(arr, n):

    for i in range(n):
        print(arr[i], end = " ")


# Function to find the original array
def findOrgArr(arr, l, r, x, n, q):

    b = [0 for i in range(n)]

    for i in range(q):

        # Decrement the element at l[i]th 
        # index by -x
        b[l[i]] += -x[i]

        # Increment the element at (r[i] + 1)th 
        # index by x if (r[i] + 1) is a valid index
        if (r[i] + 1 < n):
            b[r[i] + 1] += x[i]
    
    for i in range(n):
        
        # Prefix sum of array b
        b[i] = b[i - 1] + b[i]

    # Update the original array
    for i in range(n):
        arr[i] = arr[i] + b[i]

    printArr(arr, n)

# Driver code
arr = [20, 30, 20, 70, 100]

# Size of the array
n = len(arr)

# Queries
l = [0, 1, 3 ]
r = [2, 4, 4 ]
x = [10, 20, 30 ]

# Number of queries
q = len(l)

findOrgArr(arr, l, r, x, n, q)

# This code Is contributed by
# Mohit kumar 29

// C# implementation of above approach 
using System;

class GFG
{

// Utility function to print the 
// contents of an array 
static void printArr(int[] arr, int n) 
{ 
    for (int i = 0; i < n; i++) 
    { 
        Console.Write(arr[i] + " "); 
    } 
} 

// Function to find the original array 
static void findOrgArr(int[] arr, int[] l, 
                       int[] r, int[] x, 
                       int n, int q) 
{ 
    int[] b = new int[n];
    
    for (int i = 0; i < q; i++)
        b[i] = 0 ;

    for (int i = 0; i < q; i++)
    { 

        // Decrement the element at l[i]th 
        // index by -x 
        b[l[i]] += -x[i]; 

        // Increment the element at (r[i] + 1)th 
        // index by x if (r[i] + 1) is a valid index 
        if (r[i] + 1 < n) 
            b[r[i] + 1] += x[i]; 
    } 

    for (int i = 1; i < n; i++) 
    
        // Prefix sum of array b 
        b[i] = b[i - 1] + b[i]; 

    // Update the original array 
    for (int i = 0; i < n; i++) 
        arr[i] = arr[i] + b[i]; 

    printArr(arr, n); 
} 

// Driver code 
public static void Main()
{ 
    // Final array 
    int[] arr = { 20, 30, 20, 70, 100 }; 

    // Size of the array 
    int n = arr.Length;

    // Queries 
    int[] l = { 0, 1, 3 }; 
    int[] r = { 2, 4, 4 }; 
    int[] x = { 10, 20, 30 }; 

    // Number of queries 
    int q = l.Length;

    findOrgArr(arr, l, r, x, n, q); 
} 
}

// This code is contributed 
// by Akanksha Rai

JavaScript

<script>
 // Javascript implementation of above approach 

// Utility function to print the contents of an array 
function printArr(arr, n) 
    { 
        
        console.log(arr.join(' ')) ; 
       
    } 
    // Function to find the original array 
 function findOrgArr(arr,l,r,x,n,q) 
    { 
       let b = new Array(n) ;
        
        for (let i = 0; i < n; i++)
            b[i] = 0 ;
    
        for (let i = 0; i < q; i++)
        { 
    
            // Decrement the element at l[i]th index by -x 
            b[l[i]] += -x[i]; 
    
            // Increment the element at (r[i] + 1)th index 
            // by x if (r[i] + 1) is a valid index 
            if (r[i] + 1 < n) 
                b[r[i] + 1] += x[i]; 
        } 
        for (let i = 1; i < n; i++) 
            // Prefix sum of array b 
            b[i] = b[i - 1] + b[i]; 
    
        // Update the original array 
        for (let i = 0; i < n; i++) 
            arr[i] = arr[i] + b[i]; 
    
        printArr(arr, n); 
    } 
    
    // Driver code 
     
        // Final array 
        let arr = [ 20, 30, 20, 70, 100 ]; 
    
        // Size of the array 
        let n = arr.length ;
    
        // Queries 
        let l = [ 0, 1, 3 ]; 
        let r = [ 2, 4, 4 ]; 
        let x = [ 10, 20, 30 ]; 
    
        // Number of queries 
        let q = l.length ;
        
       // Function call
        findOrgArr(arr, l, r, x, n, q);  
        
        // This code is contributed by aarohirai2616.
 </script>

Output

10 0 -10 20 50

Complexity Analysis:

Time Complexity: O(n)
Auxiliary Space: O(n)

Find the sum of array after performing every query

ranjanmonisha233

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Article Tags :

Practice Tags :

Algorithms

Find the Initial Array from given array after range sum queries

Efficient Approach:

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