# Minimum number of operations to convert array A to array B by adding an integer into a subarray

Given two arrays A[] and B[] of length N, the task is to find the minimum number of operations in which the array A can be converted into array B where each operation consists of adding an integer K into a subarray from L to R.

Examples:

Input: A[] = {3, 7, 1, 4, 1, 2}, B[] = {3, 7, 3, 6, 3, 2}
Output: 1
Explanation:
In the above given example only one operation is required to convert from A to B: L = 3, R = 5 and K = 2
Array after the following operation:
Index 0: A = 3, B = 3
Index 1: A = 7, B = 7
Index 2: A = 1 + 2 = 3, B = 3
Index 3: A = 4 + 2 = 6, B = 6
Index 4: A = 1 + 2 = 3, B = 3
Index 5: A = 2, B = 2

Input: A[] = {1, 1, 1, 1, 1}, B[] = {1, 2, 1, 3, 1}
Output: 2
Explanation:
In the above given example only one operation is required to convert from A to B –
Operation 1: Add 1 to L = 2 to R = 2
Operation 2: Add 2 to L = 4 to R = 4

Approach: The idea is to count the consecutive elements, in array A, having equal difference with the corresponding element in the array B.

• Find the difference of the corresponding element from the array A and B:
```Difference = A[i] - B[i]
```
• If the difference of the corresponding elements is equal to 0, then the continue for checking the next index.
• Otherwise, Increase the index until the difference between consecutive elements is not equal to the previous difference of the consecutive elements
• Increment the count by 1, If all the index are iterated those are having a same difference.
• At the end, return the count as the minimum number of operations.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to find the ` `// minimum number of operations in ` `// which the array A can be converted ` `// to another array B ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find the minimum ` `// number of operations in which ` `// array A can be converted to array B ` `void` `checkArray(``int` `a[], ``int` `b[], ``int` `n) ` `{ ` `    ``int` `operations = 0; ` `    ``int` `i = 0; ` `     `  `    ``// Loop to iterate over the array ` `    ``while` `(i < n) { ` `         `  `        ``// if both elements are equal ` `        ``// then move to next element ` `        ``if` `(a[i] - b[i] == 0) { ` `            ``i++; ` `            ``continue``; ` `        ``} ` ` `  `        ``// Calculate the difference ` `        ``// between two elements ` `        ``int` `diff = a[i] - b[i]; ` `        ``i++; ` ` `  `        ``// loop while the next pair of ` `        ``// elements have same difference ` `        ``while` `(i < n && ` `           ``a[i] - b[i] == diff) { ` `            ``i++; ` `        ``} ` ` `  `        ``// Increase the number of ` `        ``// operations by 1 ` `        ``operations++; ` `    ``} ` ` `  `    ``// Print the number of ` `    ``// operations required ` `    ``cout << operations << ``"\n"``; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `a[] = { 3, 7, 1, 4, 1, 2 }; ` `    ``int` `b[] = { 3, 7, 3, 6, 3, 2 }; ` `    ``int` `size = ``sizeof``(a) / ``sizeof``(a); ` ` `  `    ``checkArray(a, b, size); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation to find the ` `// minimum number of operations in ` `// which the array A can be converted ` `// to another array B ` `class` `GFG { ` ` `  `    ``// Function to find the minimum ` `    ``// number of operations in which ` `    ``// array A can be converted to array B ` `    ``static` `void` `checkArray(``int` `a[], ``int` `b[], ``int` `n) ` `    ``{ ` `        ``int` `operations = ``0``; ` `        ``int` `i = ``0``; ` `         `  `        ``// Loop to iterate over the array ` `        ``while` `(i < n) { ` `             `  `            ``// if both elements are equal ` `            ``// then move to next element ` `            ``if` `(a[i] - b[i] == ``0``) { ` `                ``i++; ` `                ``continue``; ` `            ``} ` `     `  `            ``// Calculate the difference ` `            ``// between two elements ` `            ``int` `diff = a[i] - b[i]; ` `            ``i++; ` `     `  `            ``// loop while the next pair of ` `            ``// elements have same difference ` `            ``while` `(i < n && ` `               ``a[i] - b[i] == diff) { ` `                ``i++; ` `            ``} ` `     `  `            ``// Increase the number of ` `            ``// operations by 1 ` `            ``operations++; ` `        ``} ` `     `  `        ``// Print the number of ` `        ``// operations required ` `        ``System.out.println(operations); ` `    ``} ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `main (String[] args) ` `    ``{ ` `        ``int` `a[] = { ``3``, ``7``, ``1``, ``4``, ``1``, ``2` `}; ` `        ``int` `b[] = { ``3``, ``7``, ``3``, ``6``, ``3``, ``2` `}; ` `        ``int` `size = a.length; ` `     `  `        ``checkArray(a, b, size); ` `    ``} ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## C#

 `// C# implementation to find the ` `// minimum number of operations in ` `// which the array A can be converted ` `// to another array B ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// Function to find the minimum ` `    ``// number of operations in which ` `    ``// array A can be converted to array B ` `    ``static` `void` `checkArray(``int` `[]a, ``int` `[]b, ``int` `n) ` `    ``{ ` `        ``int` `operations = 0; ` `        ``int` `i = 0; ` `         `  `        ``// Loop to iterate over the array ` `        ``while` `(i < n) { ` `             `  `            ``// if both elements are equal ` `            ``// then move to next element ` `            ``if` `(a[i] - b[i] == 0) { ` `                ``i++; ` `                ``continue``; ` `            ``} ` `     `  `            ``// Calculate the difference ` `            ``// between two elements ` `            ``int` `diff = a[i] - b[i]; ` `            ``i++; ` `     `  `            ``// loop while the next pair of ` `            ``// elements have same difference ` `            ``while` `(i < n && ` `               ``a[i] - b[i] == diff) { ` `                ``i++; ` `            ``} ` `     `  `            ``// Increase the number of ` `            ``// operations by 1 ` `            ``operations++; ` `        ``} ` `     `  `        ``// Print the number of ` `        ``// operations required ` `        ``Console.WriteLine(operations); ` `    ``} ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `Main (``string``[] args) ` `    ``{ ` `        ``int` `[]a = { 3, 7, 1, 4, 1, 2 }; ` `        ``int` `[]b = { 3, 7, 3, 6, 3, 2 }; ` `        ``int` `size = a.Length; ` `     `  `        ``checkArray(a, b, size); ` `    ``} ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 implementation to find the ` `# minimum number of operations in ` `# which the array A can be converted ` `# to another array B ` ` `  `# Function to find the minimum ` `# number of operations in which ` `# array A can be converted to array B ` `def` `checkArray(a, b, n) : ` ` `  `    ``operations ``=` `0``; ` `    ``i ``=` `0``; ` `     `  `    ``# Loop to iterate over the array ` `    ``while` `(i < n) : ` `         `  `        ``# if both elements are equal ` `        ``# then move to next element ` `        ``if` `(a[i] ``-` `b[i] ``=``=` `0``) : ` `            ``i ``+``=` `1``; ` `            ``continue``; ` ` `  `        ``# Calculate the difference ` `        ``# between two elements ` `        ``diff ``=` `a[i] ``-` `b[i]; ` `        ``i ``+``=` `1``; ` ` `  `        ``# loop while the next pair of ` `        ``# elements have same difference ` `        ``while` `(i < n ``and` `a[i] ``-` `b[i] ``=``=` `diff) : ` `            ``i ``+``=` `1``; ` ` `  `        ``# Increase the number of ` `        ``# operations by 1 ` `        ``operations ``+``=` `1``; ` `     `  `    ``# Print the number of ` `    ``# operations required ` `    ``print``(operations); ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"` `: ` ` `  `    ``a ``=` `[ ``3``, ``7``, ``1``, ``4``, ``1``, ``2` `]; ` `    ``b ``=` `[ ``3``, ``7``, ``3``, ``6``, ``3``, ``2` `]; ` `    ``size ``=` `len``(a); ` ` `  `    ``checkArray(a, b, size); ` ` `  `# This code is contributed by AnkitRai01 `

Performance Analysis:

• Time Complexity: As in the above approach, There is only one loop which takes O(N) time in worst case. Hence the Time Complexity will be O(N).
• Auxiliary Space Complexity: As in the above approach, There is no extra space used. Hence the auxiliary space complexity will be O(1).

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Improved By : AnkitRai01