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++ 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 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 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 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 ?> |
<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> |
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++ 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 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 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 |
<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>
|
10 0 -10 20 50
Complexity Analysis:
- Time Complexity: O(n)
- Auxiliary Space: O(n)