Related Articles

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

• Difficulty Level : Easy
• Last Updated : 23 Apr, 2021

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:
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:
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 an equal difference with the corresponding element in 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 continue checking for 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, until all the indexes are iterated having the same difference.
• In 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`

## Javascript

 ``

Performance Analysis:

• Time Complexity: As in the above approach, there is only one loop that takes O(N) time in the 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).

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up