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

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 second array by adding K to a selected range [L, R] in the array. If no such number K exists, print NA

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• Create a temporary array c[] which contains the difference of the array elements, i.e.,
```ci = bi - ai
```
• Then create a vector pair for all non zero elements of array c[n] with there 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 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 temporary array c[] and 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 difference value in the first array a[n] in the given indexes [2, 4] to convert it into 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 ` `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 there index ` `    ``vector > 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.*; ` ` `  `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 there index ` `    ``Vector idxs = ``new` `Vector(); ` `    ``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 there 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 there index ` `    ``List idxs = ``new` `List(); ` `    ``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 `

Output:

```2
```

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