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[0] = 3, B[0] = 3
Index 1: A[1] = 7, B[1] = 7
Index 2: A[2] = 1 + 2 = 3, B[2] = 3
Index 3: A[3] = 4 + 2 = 6, B[3] = 6
Index 4: A[4] = 1 + 2 = 3, B[4] = 3
Index 5: A[5] = 2, B[5] = 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++

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// C++ implementation to find the
// minimum number of operations in
// which the array A can be converted
// to another array B
  
#include <bits/stdc++.h>
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[0]);
  
    checkArray(a, b, size);
  
    return 0;
}

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Java

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// 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

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C#

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// 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

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Python3

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# 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

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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