Find K such that array A can be converted into array B by adding K to a selected range [L, R]

Last Updated : 14 Mar, 2023

Given two arrays a[] and b[] of length N consisting of unique elements, the task is to find a number K (K > 0) such that the first array can be converted into the second array by adding K to a selected range [L, R] in the array. If no such number K exists, print N
Examples:

Input: a[] = {3, 7, 1, 4, 0, 2, 2}, b[] = {3, 7, 3, 6, 2, 2, 2}
Output: 2
Explanation:
Array a[] can be converted into Array b[] by adding K = 2 to range [2, 4]

Input: a[] = {3, 7, 1, 4, 0, 1, 2}, b[] = {3, 7, 3, 6, 2, 2}
Output: NA

Approach:

Create a temporary array c[] which contains the difference between the array elements, i.e.,

c_i = b_i - a_i

Then create a vector pair for all non-zero elements of the array c[n] with their index. Therefore, the vector pair will be as:

vector<c[i], o>

where c[i] is a non zero value in c[]
and i is the index of c[i]

If the index value differs by 1 and the difference value is the same, then K = difference value and [L, R] = the index range.
Hence, the array a[n] can be converted into b[n] by adding K to [a[L], a[R]].

For Example:

Given

a[n] = [3, 7, 1, 4, 0, 2, 2]
b[n] = [3, 7, 3, 6, 2, 2, 2]

So, upon creating the temporary array c[] and a vector pair:

c[n] = [0, 0, 2, 2, 2, 0, 0]
vector pair = {{2, 2}, {2, 3}, {2, 4}}

As all the index values (2, 3, 4) in the vector pair differ by 1, so they are consecutive.
And the value of the difference is the same (2).
Hence, we can simply add that different value in the first array a[n] in the given indexes [2, 4] to convert it into a second array b[n].
Hence, the required K value will be 2

Below is the implementation of the above approach:

C++

// C++ implementation of above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to Check if it is possible to
// convert a given array to another array
// by adding elements to first array
bool checkconv(int a[], int b[], int n)
{
    int c[n], flag = 0;
 
    // Create a temporary array c[]
    // which contains the difference
    // of the array elements
    for (int i = 0; i < n; i++) {
 
        c[i] = b[i] - a[i];
    }
 
    // Create a vector pair for all non zero
    // elements of array c[n] with their index
    vector<pair<int, int> > idxs;
    for (int i = 0; i < n; i++) {
        if (c[i] != 0)
            idxs.push_back(make_pair(i, c[i]));
    }
 
    // Check If the index value differs by 1
    // and the difference value is same
    for (int i = 0; i < idxs.size() - 1; i++) {
        if (idxs[i + 1].first - idxs[i].first != 1
            || idxs[i + 1].second != idxs[i].second) {
            flag = 1;
            break;
        }
    }
 
    return !flag;
}
 
// Function to calculate the value of K
int diffofarrays(int a[], int b[], int n)
{
    int c[n], ans = 0;
    for (int i = 0; i < n; i++) {
        c[i] = b[i] - a[i];
    }
    for (int i = 0; i < n; i++) {
        if (c[i] != 0) {
            ans = c[i];
            break;
        }
    }
 
    return ans;
}
 
// Driver code
int main()
{
    int A[] = { 3, 7, 1, 4, 0, 2, 2 };
    int B[] = { 3, 7, 3, 6, 2, 2, 2 };
    int arr_size = sizeof(A) / sizeof(A[0]);
 
    if (checkconv(A, B, arr_size)) {
        cout << diffofarrays(A, B, arr_size) << endl;
    }
    else
        cout << "NA" << endl;
    return 0;
}

Java

// Java implementation of above approach
import java.util.*;
 
public class GFG
{
    static class pair
    { 
        int first, second; 
        public pair(int first, int second) 
        { 
            this.first = first; 
            this.second = second; 
        } 
    }
     
// Function to Check if it is possible to
// convert a given array to another array
// by adding elements to first array
static boolean checkconv(int a[], int b[], int n)
{
    int []c = new int[n];
    int flag = 0;
 
    // Create a temporary array c[]
    // which contains the difference
    // of the array elements
    for (int i = 0; i < n; i++)
    {
 
        c[i] = b[i] - a[i];
    }
 
    // Create a vector pair for all non zero
    // elements of array c[n] with their index
    Vector<pair > idxs = new Vector<pair>();
    for (int i = 0; i < n; i++)
    {
        if (c[i] != 0)
            idxs.add(new pair(i, c[i]));
    }
 
    // Check If the index value differs by 1
    // and the difference value is same
    for (int i = 0; i < idxs.size() - 1; i++)
    {
        if (idxs.get(i + 1).first - idxs.get(i).first != 1
            || idxs.get(i + 1).second != idxs.get(i).second) 
        {
            flag = 1;
            break;
        }
    }
 
    return flag == 1 ? false:true;
}
 
// Function to calculate the value of K
static int diffofarrays(int a[], int b[], int n)
{
    int []c = new int[n];
    int ans = 0;
    for (int i = 0; i < n; i++) 
    {
        c[i] = b[i] - a[i];
    }
    for (int i = 0; i < n; i++)
    {
        if (c[i] != 0)
        {
            ans = c[i];
            break;
        }
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int A[] = { 3, 7, 1, 4, 0, 2, 2 };
    int B[] = { 3, 7, 3, 6, 2, 2, 2 };
    int arr_size = A.length;
 
    if (checkconv(A, B, arr_size))
    {
        System.out.print(diffofarrays(A, B, arr_size) +"\n");
    }
    else
        System.out.print("NA" +"\n");
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of above approach 
 
# Function to Check if it is possible to 
# convert a given array to another array 
# by adding elements to first array 
def checkconv(a, b, n) : 
 
    c = [0]*n; flag = 0; 
 
    # Create a temporary array c[] 
    # which contains the difference 
    # of the array elements 
    for i in range(n) :
        c[i] = b[i] - a[i]; 
 
    # Create a vector pair for all non zero 
    # elements of array c[n] with their index 
    idxs = []; 
    for i in range(n) :
        if (c[i] != 0) :
            idxs.append((i, c[i])); 
 
    # Check If the index value differs by 1 
    # and the difference value is same 
    for i in range(len(idxs) - 1) :
        if (idxs[i + 1][0] - idxs[i][0] != 1
            or idxs[i + 1][1] != idxs[i][1]) :
            flag = 1; 
            break; 
 
    return not flag; 
 
# Function to calculate the value of K 
def diffofarrays(a, b, n) : 
    c = [0] * n; 
    ans = 0;
     
    for i in range(n) :
        c[i] = b[i] - a[i];
         
    for i in range(n) :
        if (c[i] != 0) :
            ans = c[i];
            break;
     
    return ans; 
 
# Driver code 
if __name__ == "__main__" : 
 
    A = [ 3, 7, 1, 4, 0, 2, 2 ];
    B = [ 3, 7, 3, 6, 2, 2, 2 ];
    arr_size = len(A);
     
    if (checkconv(A, B, arr_size)) :
        print(diffofarrays(A, B, arr_size));
         
    else :
        print("NA"); 
 
# This code is contributed by AnkitRai01

C#

// C# implementation of above approach
using System;
using System.Collections.Generic;
 
class GFG
{
    class pair
    { 
        public int first, second; 
        public pair(int first, int second) 
        { 
            this.first = first; 
            this.second = second; 
        } 
    }
     
// Function to Check if it is possible to
// convert a given array to another array
// by adding elements to first array
static bool checkconv(int []a, int []b, int n)
{
    int []c = new int[n];
    int flag = 0;
 
    // Create a temporary array c[]
    // which contains the difference
    // of the array elements
    for (int i = 0; i < n; i++)
    {
 
        c[i] = b[i] - a[i];
    }
 
    // Create a vector pair for all non zero
    // elements of array c[n] with their index
    List<pair > idxs = new List<pair>();
    for (int i = 0; i < n; i++)
    {
        if (c[i] != 0)
            idxs.Add(new pair(i, c[i]));
    }
 
    // Check If the index value differs by 1
    // and the difference value is same
    for (int i = 0; i < idxs.Count - 1; i++)
    {
        if (idxs[i + 1].first - idxs[i].first != 1
            || idxs[i + 1].second != idxs[i].second) 
        {
            flag = 1;
            break;
        }
    }
 
    return flag == 1 ? false:true;
}
 
// Function to calculate the value of K
static int diffofarrays(int []a, int []b, int n)
{
    int []c = new int[n];
    int ans = 0;
    for (int i = 0; i < n; i++) 
    {
        c[i] = b[i] - a[i];
    }
    for (int i = 0; i < n; i++)
    {
        if (c[i] != 0)
        {
            ans = c[i];
            break;
        }
    }
    return ans;
}
 
// Driver code
public static void Main(String[] args)
{
    int []A = { 3, 7, 1, 4, 0, 2, 2 };
    int []B = { 3, 7, 3, 6, 2, 2, 2 };
    int arr_size = A.Length;
 
    if (checkconv(A, B, arr_size))
    {
        Console.Write(diffofarrays(A, B, arr_size) +"\n");
    }
    else
        Console.Write("NA" +"\n");
}
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
// Javascript implementation of above approach
 
// Function to Check if it is possible to
// convert a given array to another array
// by adding elements to first array
function checkconv(a, b, n) {
    let c = new Array(n);
    let flag = 0;
 
    // Create a temporary array c[]
    // which contains the difference
    // of the array elements
    for (let i = 0; i < n; i++) {
 
        c[i] = b[i] - a[i];
    }
 
    // Create a vector pair for all non zero
    // elements of array c[n] with their index
    let idxs = new Array();
 
    for (let i = 0; i < n; i++) {
        if (c[i] != 0)
            idxs.push([i, c[i]]);
    }
 
    // Check If the index value differs by 1
    // and the difference value is same
    for (let i = 0; i < idxs.length - 1; i++) {
        if (idxs[i + 1][0] - idxs[i][0] != 1
            || idxs[i + 1][1] != idxs[i][1]) {
            flag = 1;
            break;
        }
    }
 
    return !flag;
}
 
// Function to calculate the value of K
function diffofarrays(a, b, n) {
    let c = new Array(n);
    let ans = 0;
    for (let i = 0; i < n; i++) {
        c[i] = b[i] - a[i];
    }
    for (let i = 0; i < n; i++) {
        if (c[i] != 0) {
            ans = c[i];
            break;
        }
    }
 
    return ans;
}
 
// Driver code
 
let A = [3, 7, 1, 4, 0, 2, 2];
let B = [3, 7, 3, 6, 2, 2, 2];
let arr_size = A.length;
 
if (checkconv(A, B, arr_size)) {
    document.write(diffofarrays(A, B, arr_size) + "<br>");
}
else
    document.write("NA" + "<br>");
</script>