Minimum increments by index value required to obtain at least two equal Array elements

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>

Output:

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>

Output:

4

Time Complexity: O(N)
Auxiliary Space: O(1)