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Generate Array whose difference of each element with its left yields the given Array

  • Last Updated : 05 Mar, 2021
Geek Week

Given an integer N and an arr1[], of (N – 1) integers, the task is to find the sequence arr2[] of N integers in the range [1, N] such that arr1[i] = arr2[i+1] – arr2[i]. The integers in sequence arr1[] lies in range [-N, N].
Examples: 
 

Input: N = 3, arr1[] = {-2, 1} 
Output: arr2[] = {3, 1, 2} 
Explanation: 
arr2[1] – arr2[0] = (1 – 3) = -2 = arr1[0] 
arr2[2] – arr2[1] = (2 – 1) = 1 = arr1[1]
Input: N = 5, arr1 = {1, 1, 1, 1, 1} 
Output: arr2 = {1, 2, 3, 4, 5} 
Explanation: 
arr2[1] – arr2[0] = (2 – 1) = 1 = arr1[0] 
arr2[2] – arr2[1] = (3 – 2) = 1 = arr1[1] 
arr2[3] – arr2[2] = (4 – 3) = 1 = arr1[2] 
arr2[4] – arr2[3] = (5 – 4) = 1 = arr1[3] 
 

 

Approach: 
Follow the steps to solve the problem: 
 

  1. Assume the first element of arr2[] to be X.
  2. The next element will be X + arr1[0].
  3. The rest of the elements of arr2[] can be represented, w.r.t X.
  4. It is known that the sequence arr2[] can contain integers in the range [1, N]. So the minimum possible integer would be 1.
  5. The minimum number of the arr2[] can be found out in terms of X, and equate it with 1 to find the value of X.
  6. Finally using the values of X, all the other numbers in arr2[] can be found out.

Below is the implementation of the above approach: 
 



C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the sequence
void find_seq(int arr[],
              int m, int n) {
    int b[n];
    int x = 0;
 
    // initializing 1st element
    b[0] = x;
 
    // Creating sequence in
    // terms of x
    for (int i = 0;
         i < n - 1; i++) {
 
        b[i + 1] = x +
                   arr[i] + b[i];
    }
 
    int mn = n;
 
    // Finding min element
    for (int i = 0; i < n; i++)
    {
        mn = min(mn, b[i]);
    }
 
    // Finding value of x
    x = 1 - mn;
 
    // Creating original sequence
    for (int i = 0; i < n; i++) {
        b[i] += x;
    }
 
    // Output original sequence
    for (int i = 0; i < n; i++) {
        cout << b[i] << " ";
    }
    cout << endl;
}
 
// Driver function
int main()
{
    int N = 3;
    int arr[] = { -2, 1 };
 
    int M = sizeof(arr) / sizeof(int);
    find_seq(arr, M, N);
 
    return 0;
}

Java




// Java implementation of the above approach
class GFG{
     
// Function to find the sequence
static void find_seq(int arr[], int m,
                                int n)
{
    int b[] = new int[n];
    int x = 0;
 
    // Initializing 1st element
    b[0] = x;
 
    // Creating sequence in
    // terms of x
    for(int i = 0; i < n - 1; i++)
    {
       b[i + 1] = x + arr[i] + b[i];
    }
 
    int mn = n;
 
    // Finding min element
    for(int i = 0; i < n; i++)
    {
       mn = Math.min(mn, b[i]);
    }
 
    // Finding value of x
    x = 1 - mn;
 
    // Creating original sequence
    for(int i = 0; i < n; i++)
    {
       b[i] += x;
    }
 
    // Output original sequence
    for(int i = 0; i < n; i++)
    {
        System.out.print(b[i] + " ");
    }
    System.out.println();
}
     
// Driver code
public static void main (String[] args)
{
    int N = 3;
    int arr[] = new int[]{ -2, 1 };
    int M = arr.length;
     
    find_seq(arr, M, N);
}
}
 
// This code is contributed by Pratima Pandey

Python3




# Python3 program for the above approach
 
# Function to find the sequence
def find_seq(arr, m, n):
     
    b = []
    x = 0
     
    # Initializing 1st element
    b.append(x)
     
    # Creating sequence in
    # terms of x
    for i in range(n - 1):
        b.append(x + arr[i] + b[i])
         
    mn = n
     
    # Finding min element
    for i in range(n):
        mn = min(mn, b[i])
         
    # Finding value of x
    x = 1 - mn
         
    # Creating original sequence
    for i in range(n):
        b[i] += x
         
    # Output original sequence
    for i in range(n):
        print(b[i], end = ' ')
     
    print()
     
# Driver code
if __name__=='__main__':
     
    N = 3
    arr = [ -2, 1 ]
    M = len(arr)
     
    find_seq(arr, M, N)
 
# This code is contributed by rutvik_56

C#




// C# implementation of the above approach
using System;
 
class GFG{
     
// Function to find the sequence
static void find_seq(int []arr, int m,
                                int n)
{
    int []b = new int[n];
    int x = 0;
 
    // Initializing 1st element
    b[0] = x;
 
    // Creating sequence in
    // terms of x
    for(int i = 0; i < n - 1; i++)
    {
       b[i + 1] = x + arr[i] + b[i];
    }
 
    int mn = n;
 
    // Finding min element
    for(int i = 0; i < n; i++)
    {
       mn = Math.Min(mn, b[i]);
    }
 
    // Finding value of x
    x = 1 - mn;
 
    // Creating original sequence
    for(int i = 0; i < n; i++)
    {
       b[i] += x;
    }
 
    // Output original sequence
    for(int i = 0; i < n; i++)
    {
       Console.Write(b[i] + " ");
    }
    Console.WriteLine();
}
     
// Driver code
public static void Main(String[] args)
{
    int N = 3;
    int []arr = new int[]{ -2, 1 };
    int M = arr.Length;
     
    find_seq(arr, M, N);
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// Javascript program for the above approach
 
// Function to find the sequence
function find_seq(arr,m, n) {
    let b = new Array(n);
    let x = 0;
 
    // initializing 1st element
    b[0] = x;
 
    // Creating sequence in
    // terms of x
    for (let i = 0;
         i < n - 1; i++) {
 
        b[i + 1] = x +
                   arr[i] + b[i];
    }
 
    let mn = n;
 
    // Finding min element
    for (let i = 0; i < n; i++)
    {
        mn = Math.min(mn, b[i]);
    }
 
    // Finding value of x
    x = 1 - mn;
 
    // Creating original sequence
    for (let i = 0; i < n; i++) {
        b[i] += x;
    }
 
    // Output original sequence
    for (let i = 0; i < n; i++) {
        document.write(b[i] + " ");
    }
    document.write("<br>");
}
 
// Driver function
 
    let N = 3;
    let arr = [ -2, 1 ];
 
    let M = arr.length;
    find_seq(arr, M, N);
 
  
// This code is contributed by Mayank Tyagi
 
</script>
Output: 
3 1 2

 

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

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