Given an array arr[] of length N, where arr is derived from an array nums[] which is lost. Array arr[] is derived as:
arr[i] = (nums[0] + nums[1] + … + nums[i]) + (nums[i] + nums[i+1] + … + nums[N-1]).
The task is to find nums[] array of length N.
Examples:
Input: N = 4, arr[] = {9, 10, 11, 10}
Output: {1, 2, 3, 2}
Explanation: If nums[] = {1, 2, 3, 2}, then according to above definition
arr[0] = (nums[0]) + (nums[0] + nums[1] + nums[2] + nums[3]) = 1 + 1 + 2 + 3 + 2 = 9
arr[1] = (nums[0] + nums[1]) + (nums[1] + nums[2] + nums[3]) = 1 + 2 + 2 + 3 + 2 = 10
arr[2] = (nums[0] + nums[1] + nums[2]) + (nums[2] + nums[3]) = 1 + 2 + 3 + 3 + 2 = 11
arr[3] = (nums[0] + nums[1] + nums[2] + nums[3]) + (nums[3]) = 1 + 2 + 3 + 2 + 2 = 10Input: N = 2, arr[] = [25, 20]
Output: [10, 5]
Approach: Follow the below idea to solve the problem:
Suppose nums[] contains [a1, a2, a3, …, aN]
Then, sum = a1 + a2 + a3 + . . . + aN.
We are given
b1 = a1 + a1 + a2 + . . . + aN = a1 + sum …..(1)
Similarly,
b2 = a1 + a2 + a2 + . . . + aN = a2 + sum …..(2)
. . . (so on) and in last
b1 = a1 + a2 + a3 + . . . + aN + aN = aN + sum …..(N)
where [b1, b2, b3 , . . ., bN] are elements of arr[] and,
total = b1 + b2 + b3 + . . . + bNAdding all equation (1) + (2) + (3) + …. + (N) we will get
b1 + b2 + b3 + . . . + bN = (a1 + sum) + (a2 + sum) + . . . + (aN + sum)
total = (a1 + a1 + a2 + . . . + aN) + (N * sum)
total = (sum) + (N * sum)
total = (N + 1) * sumNow find the value of sum variable after that simply:
a1 = (b1 – sum), a2 = (b2 – sum), . . ., aN = (bN – sum)
Using the above idea follow the below steps to implement the code:
- First of all, try to store the sum of elements of arr[] in a variable let’s say total
- Using the formula (N + 1) * sum = total, we will get the value of variable sum which denotes the sum of elements present in the nums[] array.
- At last traverse N times to find nums[0] = arr[0] – sum, nums[1] = arr[1] – sum and so on.
- Return the array and print it.
Below is the implementation of the above approach:
// C++ Algorithm for the above approach #include <iostream> #include <vector> using namespace std;
// Function to find the original // array nums[] vector< int > findOrgArray(vector< int > arr, int N)
{ // Total variable stores the sum of
// elements of arr[]
int total = 0;
for ( int val : arr)
total += val;
// Sum variable stores the sum of
// elements of nums[]
int sum = (total / (N + 1));
vector< int > v;
// Traversing to find the elements
// of nums[]
for ( int i = 0; i < N; i++) {
int val = arr[i] - sum;
v.push_back(val);
}
// Returning nums[]
return v;
} int main()
{ int N = 4;
vector< int > arr = { 9, 10, 11, 10 };
vector< int > v = findOrgArray(arr, N);
for ( auto val : v)
cout << val << " " ;
return 0;
} |
// Java algorithm of the above approach import java.util.*;
class GFG {
// Driver Code
public static void main(String[] args)
{
int N = 4 ;
int [] arr = { 9 , 10 , 11 , 10 };
List<Integer> nums = findOrgArray(arr, N);
for ( int x : nums)
System.out.print(x + " " );
}
// Function to find the original
// array nums[]
public static List<Integer> findOrgArray( int [] arr,
int N)
{
// Total variable stores the sum of
// elements of arr[]
int total = 0 ;
for ( int val : arr)
total += val;
// Sum variable stores the sum of
// elements of nums[]
int sum = (total / (N + 1 ));
List<Integer> nums = new ArrayList<>();
// Traversing to find the elements
// of nums[]
for ( int i = 0 ; i < N; i++) {
int val = arr[i] - sum;
nums.add(val);
}
// Returning nums[]
return nums;
}
} |
# python3 Algorithm for the above approach # Function to find the original # array nums[] def findOrgArray(arr, N) :
# Total variable stores the sum of
# elements of arr[]
total = 0
for i in arr :
total + = i
# Sum variable stores the sum of
# elements of nums[]
sum = int (total / (N + 1 ));
v = []
# Traversing to find the elements
# of nums[]
for i in range (N) :
val = arr[i] - sum
v.append(val)
# Returning nums[]
return v
# Driver Code if __name__ = = "__main__" :
N = 4
arr = [ 9 , 10 , 11 , 10 ]
v = findOrgArray(arr, N)
for val in v :
print (val,end = ' ' )
# this code is contributed by aditya942003patil |
// C# program to implement // the above approach using System;
using System.Collections.Generic;
public class GFG
{ // Function to find the original
// array nums[]
public static List< int > findOrgArray( int [] arr,
int N)
{
// Total variable stores the sum of
// elements of arr[]
int total = 0;
//for (int x = 0; x < arr.count; x++)
foreach ( int val in arr)
total += val;
// Sum variable stores the sum of
// elements of nums[]
int sum = (total / (N + 1));
List< int > nums = new List< int >();
// Traversing to find the elements
// of nums[]
for ( int i = 0; i < N; i++) {
int val = arr[i] - sum;
nums.Add(val);
}
// Returning nums[]
return nums;
}
// Driver Code
public static void Main(String []args)
{
int N = 4;
int [] arr = { 9, 10, 11, 10 };
List< int > nums = findOrgArray(arr, N);
for ( int x = 0; x < nums.Count; x++)
Console.Write(nums[x] + " " );
}
} // This code is contributed by sanjoy_62. |
<script> // Function to find the original // array nums[] function findOrgArray(arr, N)
{ // Total variable stores the sum of
// elements of arr[]
let total = 0;
for (let i = 0; i < N; i++)
total += arr[i];
// Sum variable stores the sum of
// elements of nums[]
let sum = (total / (N + 1));
let v= new Array(N);
// Traversing to find the elements
// of nums[]
for (let i = 0; i < N; i++) {
v[i] = arr[i] - sum;
}
// Returning nums[]
return v;
} let N = 4;
let arr = [ 9, 10, 11, 10 ];
let v = findOrgArray(arr, N);
for (let i = 0; i < N; i++)
document.write(v[i]+ " " );
// This code is contributed by satwik4409.
</script>
|
1 2 3 2
Time Complexity: O(N)
Auxiliary Space: O(N), to further reduce it to O(1), store the value in the same given array arr[] rather than storing it in a new array.
Another Approach:
- Initialize a variable named “total” to 0.
-
Traverse the input array “arr” using a range-based for loop.
a. For each element “val” in “arr”, add “val” to the “total” variable. - Compute the sum of the original array “nums” using the formula: sum = total / (N + 1).
-
Traverse the input array “arr” again using a for loop with index “i” from 0 to N-1.
a. For each element in “arr”, subtract “sum” from it and store the result back into “arr[i]”. This effectively undoes the modification made to “arr” and recovers the original array “nums”.
Below is the implementation of the above approach:
#include <iostream> #include <vector> using namespace std;
// Function to find the original array nums[] void findOrgArray(vector< int >& arr, int N)
{ // Total variable stores the sum of elements of arr[]
int total = 0;
for ( int val : arr)
total += val;
// Sum variable stores the sum of elements of nums[]
int sum = (total / (N + 1));
// Traversing to find the elements of nums[]
for ( int i = 0; i < N; i++) {
arr[i] = arr[i] - sum;
}
} int main()
{ int N = 4;
vector< int > arr = { 9, 10, 11, 10 };
findOrgArray(arr, N);
for ( auto val : arr)
cout << val << " " ;
return 0;
} |
import java.util.*;
import java.io.*;
public class GFG {
// Function to find the original array nums[]
static void findOrgArray(List<Integer> arr, int N) {
// Total variable stores the sum of elements of arr[]
int total = 0 ;
for ( int val : arr) {
total += val;
}
// Sum variable stores the sum of elements of nums[]
int sum = total / (N + 1 );
// Traversing to find the elements of nums[]
for ( int i = 0 ; i < N; i++) {
arr.set(i, arr.get(i) - sum);
}
}
// Driver Code
public static void main(String[] args) {
int N = 4 ;
List<Integer> arr = new ArrayList<>();
arr.add( 9 );
arr.add( 10 );
arr.add( 11 );
arr.add( 10 );
findOrgArray(arr, N);
for ( int val : arr) {
System.out.print(val + " " );
}
}
} |
#Function to find the original array nums[] def find_org_array(arr, N):
total = 0 #Total variable stores the sum of elements of arr[]
for val in arr:
total + = val
#Sum variable stores the sum of elements of nums[]
sum = int (total / (N + 1 ))
#Traversing to find the elements of nums[]
for i in range (N):
arr[i] - = sum
return arr
#Driver Code arr = [ 9 , 10 , 11 , 10 ]
n = 4
#function call arr = find_org_array(arr,n)
for val in arr:
print (val,end = " " )
|
using System;
using System.Collections.Generic;
class Program
{ // Function to find the original array nums[]
static void FindOrgArray(List< int > arr, int N)
{
// Total variable stores the sum of elements of arr[]
int total = 0;
foreach ( int val in arr)
total += val;
// Sum variable stores the sum of elements of nums[]
int sum = total / (N + 1);
// Traversing to find the elements of nums[]
for ( int i = 0; i < N; i++)
{
arr[i] = arr[i] - sum;
}
}
static void Main()
{
int N = 4;
List< int > arr = new List< int > { 9, 10, 11, 10 };
FindOrgArray(arr, N);
foreach ( var val in arr)
Console.Write(val + " " );
}
} |
// Function to find the original array nums[] function findOrgArray(arr, N) {
// Total variable stores the sum of elements of arr[]
let total = 0;
for (let val of arr) {
total += val;
}
// Sum variable stores the sum of elements of nums[]
let sum = Math.floor(total / (N + 1));
// Traversing to find the elements of nums[]
for (let i = 0; i < N; i++) {
arr[i] = arr[i] - sum;
}
} // Driver code let N = 4; let arr = [9, 10, 11, 10]; findOrgArray(arr, N); for (let val of arr) {
console.log(val + " " );
} |
1 2 3 2
Time Complexity: O(n)
Auxiliary Space: O(1)