Given a strictly decreasing array arr[] consisting of N integers, the task is to find the minimum number of operations required to make at least two array elements equal, where each operation involves increasing every array element by its index value.
Examples:
Input: arr[] = {6, 5, 1}
Output: 1
Explanation:
{6 + 1, 5 + 2, 1 + 3} = {7, 7, 4}
Input: arr[] = {12, 8, 4}
Output: 4
Explanation:
Step 1 : {12 + 1, 8 + 2, 4 + 3} = {13, 10, 7}
Step 2 : {13 + 1, 10 + 2, 7 + 3} = {14, 12, 10}
Step 3 : {15, 14, 13}
Step 4 : {16, 16, 16}
Naive approach: Follow the below steps to solve the problem:
- Check if the array already has at least two equal elements or not. If found to be true, print 0.
- Otherwise, keep updating the array by increasing each array element by its index value and increase count. Check if array has two equal elements or not.
- Print count once the array is found to be containing at least two equal elements.
Below is the implementation of the above approach:
C++
// C++ Program to implement // the above approach #include <bits/stdc++.h> using namespace std; // Function to update every element // adding to it its index value void update( int arr[], int N) { for ( int i = 0; i < N; i++) { arr[i] += (i + 1); } } // Function to check if at least // two elements are equal or not bool check( int arr[], int N) { bool f = 0; for ( int i = 0; i < N; i++) { // Count the frequency of arr[i] int count = 0; for ( int j = 0; j < N; j++) { if (arr[i] == arr[j]) { count++; } } if (count >= 2) { f = 1; break ; } } if (f == 1) return true ; else return false ; } // Function to calculate the number // of increment operations required void incrementCount( int arr[], int N) { // Stores the minimum number of steps int min = 0; while (check(arr, N) != true ) { update(arr, N); min++; } cout << min; } // Driver Code int main() { int N = 3; int arr[N] = { 12, 8, 4 }; incrementCount(arr, N); return 0; } |
Java
// Java program to implement // the above approach import java.util.*; class GFG{ // Function to update every element // adding to it its index value static void update( int arr[], int N) { for ( int i = 0 ; i < N; i++) { arr[i] += (i + 1 ); } } // Function to check if at least // two elements are equal or not static boolean check( int arr[], int N) { int f = 0 ; for ( int i = 0 ; i < N; i++) { // Count the frequency of arr[i] int count = 0 ; for ( int j = 0 ; j < N; j++) { if (arr[i] == arr[j]) { count++; } } if (count >= 2 ) { f = 1 ; break ; } } if (f == 1 ) return true ; else return false ; } // Function to calculate the number // of increment operations required static void incrementCount( int arr[], int N) { // Stores the minimum number of steps int min = 0 ; while (check(arr, N) != true ) { update(arr, N); min++; } System.out.println(min); } // Driver code public static void main (String[] args) { int N = 3 ; int arr[] = { 12 , 8 , 4 }; incrementCount(arr, N); } } // This code is contributed by offbeat |
Python3
# Python3 program to implement # the above approach # Function to update every element # adding to it its index value def update(arr, N): for i in range (N): arr[i] + = (i + 1 ); # Function to check if at least # two elements are equal or not def check(arr, N): f = 0 ; for i in range (N): # Count the frequency of arr[i] count = 0 ; for j in range (N): if (arr[i] = = arr[j]): count + = 1 ; if (count > = 2 ): f = 1 ; break ; if (f = = 1 ): return True ; else : return False ; # Function to calculate the number # of increment operations required def incrementCount(arr, N): # Stores the minimum number of steps min = 0 ; while (check(arr, N) ! = True ): update(arr, N); min + = 1 ; print ( min ); # Driver code if __name__ = = '__main__' : N = 3 ; arr = [ 12 , 8 , 4 ]; incrementCount(arr, N); # This code is contributed by 29AjayKumar |
C#
// C# program to implement // the above approach using System; class GFG{ // Function to update every element // adding to it its index value static void update( int []arr, int N) { for ( int i = 0; i < N; i++) { arr[i] += (i + 1); } } // Function to check if at least // two elements are equal or not static bool check( int []arr, int N) { int f = 0; for ( int i = 0; i < N; i++) { // Count the frequency of arr[i] int count = 0; for ( int j = 0; j < N; j++) { if (arr[i] == arr[j]) { count++; } } if (count >= 2) { f = 1; break ; } } if (f == 1) return true ; else return false ; } // Function to calculate the number // of increment operations required static void incrementCount( int []arr, int N) { // Stores the minimum number of steps int min = 0; while (check(arr, N) != true ) { update(arr, N); min++; } Console.WriteLine(min); } // Driver code public static void Main(String[] args) { int N = 3; int []arr = { 12, 8, 4 }; incrementCount(arr, N); } } // This code is contributed by Amit Katiyar |
Javascript
<script> // Java Script program to implement // the above approach // Function to update every element // adding to it its index value function update(arr,N) { for (let i = 0; i < N; i++) { arr[i] += (i + 1); } } // Function to check if at least // two elements are equal or not function check(arr,N) { let f = 0; for (let i = 0; i < N; i++) { // Count the frequency of arr[i] let count = 0; for (let j = 0; j < N; j++) { if (arr[i] == arr[j]) { count++; } } if (count >= 2) { f = 1; break ; } } if (f == 1) return true ; else return false ; } // Function to calculate the number // of increment operations required function incrementCount(arr,N) { // Stores the minimum number of steps let min = 0; while (check(arr, N) != true ) { update(arr, N); min++; } document.write(min); } // Driver code let N = 3; let arr= [12, 8, 4 ]; incrementCount(arr, N); //contributed by 171fa07058 </script> |
4
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized by observing that by the given operation, the difference between any two adjacent elements reduces by 1 as the array is decreasing. Therefore, the minimum number of operations required is equal to the minimum difference between any two adjacent elements.
Below is the implementation of the above approach:
C++
// C++ Program to implement // the above approach #include <bits/stdc++.h> using namespace std; // Function to calculate the minimum // number of steps required void incrementCount( int arr[], int N) { // Stores minimum difference int mini = arr[0] - arr[1]; for ( int i = 2; i < N; i++) { mini = min(mini, arr[i - 1] - arr[i]); } cout << mini; } // Driver Code int main() { int N = 3; int arr[N] = { 12, 8, 4 }; incrementCount(arr, N); return 0; } |
Java
// Java program to implement // the above approach import java.util.*; class GFG{ // Function to calculate the minimum // number of steps required static void incrementCount( int arr[], int N) { // Stores minimum difference int mini = arr[ 0 ] - arr[ 1 ]; for ( int i = 2 ; i < N; i++) { mini = Math.min(mini, arr[i - 1 ] - arr[i]); } System.out.println(mini); } // Driver code public static void main (String[] args) { int N = 3 ; int arr[] = { 12 , 8 , 4 }; incrementCount(arr, N); } } // This code is contributed by offbeat |
Python3
# Python3 program to implement # the above approach # Function to calculate the minimum # number of steps required def incrementCount(arr, N): # Stores minimum difference mini = arr[ 0 ] - arr[ 1 ] for i in range ( 2 , N): mini = min (mini, arr[i - 1 ] - arr[i]) print (mini) # Driver Code N = 3 arr = [ 12 , 8 , 4 ] # Function call incrementCount(arr, N) # This code is contributed by Shivam Singh |
C#
// C# program to implement // the above approach using System; class GFG{ // Function to calculate the minimum // number of steps required static void incrementCount( int []arr, int N) { // Stores minimum difference int mini = arr[0] - arr[1]; for ( int i = 2; i < N; i++) { mini = Math.Min(mini, arr[i - 1] - arr[i]); } Console.WriteLine(mini); } // Driver code public static void Main(String[] args) { int N = 3; int []arr = { 12, 8, 4 }; incrementCount(arr, N); } } // This code is contributed by 29AjayKumar |
Javascript
<script> // JavaScript program for the above approach // Function to calculate the minimum // number of steps required function incrementCount(arr, N) { // Stores minimum difference let mini = arr[0] - arr[1]; for (let i = 2; i < N; i++) { mini = Math.min(mini, arr[i - 1] - arr[i]); } document.write(mini); } // Driver Code let N = 3; let arr = [ 12, 8, 4 ]; incrementCount(arr, N); </script> |
4
Time Complexity: O(N)
Auxiliary Space: O(1)
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