Minimize steps required to make all array elements same by adding 1, 2 or 5
Last Updated :
15 Jul, 2025
Given an array arr[] of size N, the task is to count the minimum number of steps required to make all the array elements the same by adding 1, 2, or 5 to exactly (N - 1) elements of the array at each step.
Examples:
Input: N = 4, arr[] = {2, 2, 3, 7}
Output: 2
Explanation:
Step 1: {2, 2, 3, 7} -> {3, 3, 3, 8}
Step 2: {3, 3, 3, 8} -> {8, 8, 8, 8}
Input: N = 3, arr[] = {10, 7, 12}
Output: 3
Naive Approach: The simplest approach is to try all possible combinations recursively of adding numbers 1, 2, and 5 such that all the elements become the same and calculate the number of steps required for all such combinations. Finally, print the minimum of them as the required answer.
Time Complexity: O(MM), where M is the maximum element present in the array.
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized by the following observations:
- Adding a number K to all indices except one (say index X) is the same as removing K from the value at index X.
- This reduces the bound to search for the final array element to lower than equal to the minimum value present in the given array.
- Let the minimum value be A, then the final value after optimal operations can either be A, A - 1, or A - 2.
- The reason A - 3 and so on are not considered in the calculation is because A + 1 takes 2 steps to reach there (-1, -2 ), A + 2 requires one step in reaching A - 3 (- 5) but can easily reach A requires in a single step(A+1 requires 1 step(-1) to reach A and A + 2 requires 1 step(-2) to reach A).
- Also, A + 3 requires 2 steps (-5, -1) to reach A - 3 and 2 steps to reach A again (-1, -2).
- Therefore, A - 3 or any lower bases are not needed to be considered.
Therefore, the idea is to find the count of operations required to reduce all the array elements to their minimum element(say minE), minE - 1, and minE - 2 by subtracting 1, 2, and 5. Print the minimum among the above three operations.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to calculate the minimum
// number of steps
int calculate_steps(int arr[], int n,
int minimum)
{
// count stores number of operations
// required to make all elements
// equal to minimum value
int count = 0;
// Remark, the array should remain
// unchanged for further calculations
// with different minimum
for (int i = 0; i < n; i++) {
// Storing the current value of
// arr[i] in val
int val = arr[i];
if (arr[i] > minimum) {
// Finds how much extra amount
// is to be removed
arr[i] = arr[i] - minimum;
// Subtract the maximum number
// of 5 and stores remaining
count += arr[i] / 5;
arr[i] = arr[i] % 5;
// Subtract the maximum number
// of 2 and stores remaining
count += arr[i] / 2;
arr[i] = arr[i] % 2;
if (arr[i]) {
count++;
}
}
// Restores the actual value
// of arr[i]
arr[i] = val;
}
// Return the count
return count;
}
// Function to find the minimum number
// of steps to make array elements same
int solve(int arr[], int n)
{
// Sort the array in descending order
sort(arr, arr + n, greater<int>());
// Stores the minimum array element
int minimum = arr[n - 1];
int count1 = 0, count2 = 0, count3 = 0;
// Stores the operations required
// to make array elements equal to minimum
count1 = calculate_steps(arr, n, minimum);
// Stores the operations required
// to make array elements equal to minimum - 1
count2 = calculate_steps(arr, n, minimum - 1);
// Stores the operations required
// to make array elements equal to minimum - 2
count3 = calculate_steps(arr, n, minimum - 2);
// Return minimum of the three counts
return min(count1, min(count2, count3));
}
// Driver Code
int main()
{
int arr[] = { 3, 6, 6 };
int N = sizeof(arr) / sizeof(arr[0]);
cout << solve(arr, N);
}
// Java program for the above approach
import java.util.*;
class GFG{
// Function to calculate the minimum
// number of steps
static int calculate_steps(Integer arr[],
int n, int minimum)
{
// count stores number of operations
// required to make all elements
// equal to minimum value
int count = 0;
// Remark, the array should remain
// unchanged for further calculations
// with different minimum
for (int i = 0; i < n; i++)
{
// Storing the current value of
// arr[i] in val
int val = arr[i];
if (arr[i] > minimum)
{
// Finds how much extra amount
// is to be removed
arr[i] = arr[i] - minimum;
// Subtract the maximum number
// of 5 and stores remaining
count += arr[i] / 5;
arr[i] = arr[i] % 5;
// Subtract the maximum number
// of 2 and stores remaining
count += arr[i] / 2;
arr[i] = arr[i] % 2;
if (arr[i] > 0)
{
count++;
}
}
// Restores the actual value
// of arr[i]
arr[i] = val;
}
// Return the count
return count;
}
// Function to find the minimum number
// of steps to make array elements same
static int solve(Integer arr[], int n)
{
// Sort the array in descending order
Arrays.sort(arr, Collections.reverseOrder());
// Stores the minimum array element
int minimum = arr[n - 1];
int count1 = 0, count2 = 0, count3 = 0;
// Stores the operations required
// to make array elements equal
// to minimum
count1 = calculate_steps(arr, n,
minimum);
// Stores the operations required
// to make array elements equal to
// minimum - 1
count2 = calculate_steps(arr, n,
minimum - 1);
// Stores the operations required
// to make array elements equal to
// minimum - 2
count3 = calculate_steps(arr, n,
minimum - 2);
// Return minimum of the three counts
return Math.min(count1, Math.min(count2,
count3));
}
// Driver Code
public static void main(String[] args)
{
Integer arr[] = {3, 6, 6};
int N = arr.length;
System.out.print(solve(arr, N));
}
}
// This code is contributed by Rajput-Ji
# Python3 program for the above approach
# Function to calculate the minimum
# number of steps
def calculate_steps(arr, n, minimum):
# count stores number of operations
# required to make all elements
# equal to minimum value
count = 0
# Remark, the array should remain
# unchanged for further calculations
# with different minimum
for i in range(n):
# Storing the current value of
# arr[i] in val
val = arr[i]
if (arr[i] > minimum):
# Finds how much extra amount
# is to be removed
arr[i] = arr[i] - minimum
# Subtract the maximum number
# of 5 and stores remaining
count += arr[i] // 5
arr[i] = arr[i] % 5
# Subtract the maximum number
# of 2 and stores remaining
count += arr[i] // 2
arr[i] = arr[i] % 2
if (arr[i]):
count += 1
# Restores the actual value
# of arr[i]
arr[i] = val
# Return the count
return count
# Function to find the minimum number
# of steps to make array elements same
def solve(arr, n):
# Sort the array in descending order
arr = sorted(arr)
arr = arr[::-1]
# Stores the minimum array element
minimum = arr[n - 1]
count1 = 0
count2 = 0
count3 = 0
# Stores the operations required
# to make array elements equal to minimum
count1 = calculate_steps(arr, n, minimum)
# Stores the operations required
# to make array elements equal to minimum - 1
count2 = calculate_steps(arr, n, minimum - 1)
# Stores the operations required
# to make array elements equal to minimum - 2
count3 = calculate_steps(arr, n, minimum - 2)
# Return minimum of the three counts
return min(count1, min(count2, count3))
# Driver Code
if __name__ == '__main__':
arr = [ 3, 6, 6 ]
N = len(arr)
print(solve(arr, N))
# This code is contributed by mohit kumar 29
// C# program for the above approach
using System;
public class GFG{
// Function to calculate the minimum
// number of steps
static int calculate_steps(int []arr, int n,
int minimum)
{
// count stores number of operations
// required to make all elements
// equal to minimum value
int count = 0;
// Remark, the array should remain
// unchanged for further calculations
// with different minimum
for (int i = 0; i < n; i++) {
// Storing the current value of
// arr[i] in val
int val = arr[i];
if (arr[i] > minimum) {
// Finds how much extra amount
// is to be removed
arr[i] = arr[i] - minimum;
// Subtract the maximum number
// of 5 and stores remaining
count += arr[i] / 5;
arr[i] = arr[i] % 5;
// Subtract the maximum number
// of 2 and stores remaining
count += arr[i] / 2;
arr[i] = arr[i] % 2;
if (arr[i]>0) {
count++;
}
}
// Restores the actual value
// of arr[i]
arr[i] = val;
}
// Return the count
return count;
}
// Function to find the minimum number
// of steps to make array elements same
static int solve(int []arr, int n)
{
// Sort the array in descending order
Array.Sort(arr);
Array.Reverse(arr);
// Stores the minimum array element
int minimum = arr[n - 1];
int count1 = 0, count2 = 0, count3 = 0;
// Stores the operations required
// to make array elements equal to minimum
count1 = calculate_steps(arr, n, minimum);
// Stores the operations required
// to make array elements equal to minimum - 1
count2 = calculate_steps(arr, n, minimum - 1);
// Stores the operations required
// to make array elements equal to minimum - 2
count3 = calculate_steps(arr, n, minimum - 2);
// Return minimum of the three counts
return Math.Min(count1, Math.Min(count2, count3));
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 3, 6, 6 };
int N = arr.Length;
Console.Write(solve(arr, N));
}
}
// This code contributed by Rajput-Ji
<script>
// JavaScript program for the above approach
// Function to calculate the minimum
// number of steps
function calculate_steps(arr, n, minimum) {
// count stores number of operations
// required to make all elements
// equal to minimum value
var count = 0;
// Remark, the array should remain
// unchanged for further calculations
// with different minimum
for (var i = 0; i < n; i++) {
// Storing the current value of
// arr[i] in val
var val = arr[i];
if (arr[i] > minimum) {
// Finds how much extra amount
// is to be removed
arr[i] = arr[i] - minimum;
// Subtract the maximum number
// of 5 and stores remaining
count += parseInt(arr[i] / 5);
arr[i] = arr[i] % 5;
// Subtract the maximum number
// of 2 and stores remaining
count += parseInt(arr[i] / 2);
arr[i] = arr[i] % 2;
if (arr[i]) {
count++;
}
}
// Restores the actual value
// of arr[i]
arr[i] = val;
}
// Return the count
return count;
}
// Function to find the minimum number
// of steps to make array elements same
function solve(arr, n) {
// Sort the array in descending order
arr.sort((a, b) => b - a);
// Stores the minimum array element
var minimum = arr[n - 1];
var count1 = 0,
count2 = 0,
count3 = 0;
// Stores the operations required
// to make array elements equal to minimum
count1 = calculate_steps(arr, n, minimum);
// Stores the operations required
// to make array elements equal to minimum - 1
count2 = calculate_steps(arr, n, minimum - 1);
// Stores the operations required
// to make array elements equal to minimum - 2
count3 = calculate_steps(arr, n, minimum - 2);
// Return minimum of the three counts
return Math.min(count1, Math.min(count2, count3));
}
// Driver Code
var arr = [3, 6, 6];
var N = arr.length;
document.write(solve(arr, N));
</script>
Time Complexity: O(N), where N is the size of the given array.
Auxiliary Space: O(1)
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