Given an array arr[]. The task is to minimize the number of operations required to make all the elements in arr[] equal. It is allowed to replace any element in arr[] with any other element almost once. Find the minimum number of operations required to do so, in one operation take any suffix of arr[] and increment/decrement values in that suffix by 1.
Examples
Input: arr[] = {-1, 0, 2}
Output: 1
Explanation: Following are the operations done to make all the elements in arr[] to be equal.
Initially, change the last element of array to 0, so arr[] = {-1, 0, 0}
Now, using the operation once on the suffix starting at arr2, which means arr2 and arr3 are decreased by 1 . Thus, making all elements of array -1.
Hence, the number of operations is 1.Input: arr[] = {-3, -5, -2, 1 }
Output : 4
Approach: This problem is implementation-based. Follow the steps below to solve the given problem.
- Since, it is not required to do any operation on suffix starting at arr1, since that can change all integers in the array.
- So, the only way to make arri equal to arri-1 is to perform an operation on suffix starting at ai, abs(ai−ai-1) times.
- Now, the optimal way to initially change a value in the array is to minimize the operations.
- In order to make arr1 equal to arr2, minimum operations are decreased by abs (arr2 - arr1).
- Similarly, for making arrn equal to arrn-1, operation = abs(arrn - arrn-1).
- For left elements, changing any element arri , affects both abs(ai−ai-1) and abs(ai+1−ai).
- Also, observe this important fact that this value is minimized, when ai is between ai-1 and ai+1, inclusive.
- Thus, number of operations is decreased from abs(ai−ai-1)+abs(ai+1−ai) to abs(ai+1−ai-1).
- Return the final answer.
Below is the implementation of the above approach:
// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
void findMinOperations(vector<int> ar, int n)
{
// Initializing vector to avoid overflows
vector<int> arr(n + 5);
for (int i = 1; i <= n; i++) {
arr[i] = ar[i - 1];
}
int result = 0;
// Calculating minimum operations to be
// performed on initial array
for (int i = 2; i <= n; i++) {
result += abs(arr[i] - arr[i - 1]);
}
// Way to change a value to make
// a1 equal to a2 or a(n)
// equal to a(n-1)
int max_operations
= max(abs(arr[1] - arr[2]),
abs(arr[n] - arr[n - 1]));
for (int i = 2; i < n; i++) {
// For the rest of elements
// taking the max of
// operations already done +
// the ones performed here
max_operations
= max(
max_operations,
abs(arr[i] - arr[i - 1])
+ abs(arr[i + 1] - arr[i])
- abs(arr[i + 1] - arr[i - 1]));
}
// Print the final result
cout << result - max_operations << "\n";
}
// Driver Code
int main()
{
int N = 3;
vector<int> arr = { -1, 0, 2 };
findMinOperations(arr, N);
return 0;
}
// Java program for above approach
class GFG {
static void findMinOperations(int[] ar, int n) {
// Initializing vector to avoid overflows
int[] arr = new int[n + 5];
for (int i = 1; i <= n; i++) {
arr[i] = ar[i - 1];
}
int result = 0;
// Calculating minimum operations to be
// performed on initial array
for (int i = 2; i <= n; i++) {
result += Math.abs(arr[i] - arr[i - 1]);
}
// Way to change a value to make
// a1 equal to a2 or a(n)
// equal to a(n-1)
int max_operations = Math.max(Math.abs(arr[1] - arr[2]),
Math.abs(arr[n] - arr[n - 1]));
for (int i = 2; i < n; i++) {
// For the rest of elements
// taking the max of
// operations already done +
// the ones performed here
max_operations = Math.max(
max_operations,
Math.abs(arr[i] - arr[i - 1])
+ Math.abs(arr[i + 1] - arr[i])
- Math.abs(arr[i + 1] - arr[i - 1]));
}
// Print the final result
System.out.println(result - max_operations);
}
// Driver Code
public static void main(String args[]) {
int N = 3;
int[] arr = { -1, 0, 2 };
findMinOperations(arr, N);
}
}
// This code is contributed by saurabh_jaiswal.
# Python code for the above approach
def findMinOperations(ar, n):
# Initializing vector to avoid overflows
arr = [0] * (n + 5)
for i in range(1, n + 1):
arr[i] = ar[i - 1]
result = 0
# Calculating minimum operations to be
# performed on initial array
for i in range(2, n + 1):
result += abs(arr[i] - arr[i - 1])
# Way to change a value to make
# a1 equal to a2 or a(n)
# equal to a(n-1)
max_operations = max(abs(arr[1] - arr[2]), abs(arr[n] - arr[n - 1]))
for i in range(2, n):
# For the rest of elements
# taking the max of
# operations already done +
# the ones performed here
max_operations = max(
max_operations,
abs(arr[i] - arr[i - 1])
+ abs(arr[i + 1] - arr[i])
- abs(arr[i + 1] - arr[i - 1]))
# Print the final result
print((result - max_operations))
# Driver Code
N = 3
arr = [-1, 0, 2]
findMinOperations(arr, N)
# This code is contributed by gfgking
// C# program to implement
// the above approach
using System;
class GFG
{
static void findMinOperations(int[] ar, int n) {
// Initializing vector to avoid overflows
int[] arr = new int[n + 5];
for (int i = 1; i <= n; i++) {
arr[i] = ar[i - 1];
}
int result = 0;
// Calculating minimum operations to be
// performed on initial array
for (int i = 2; i <= n; i++) {
result += Math.Abs(arr[i] - arr[i - 1]);
}
// Way to change a value to make
// a1 equal to a2 or a(n)
// equal to a(n-1)
int max_operations = Math.Max(Math.Abs(arr[1] - arr[2]),
Math.Abs(arr[n] - arr[n - 1]));
for (int i = 2; i < n; i++) {
// For the rest of elements
// taking the max of
// operations already done +
// the ones performed here
max_operations = Math.Max(
max_operations,
Math.Abs(arr[i] - arr[i - 1])
+ Math.Abs(arr[i + 1] - arr[i])
- Math.Abs(arr[i + 1] - arr[i - 1]));
}
// Print the final result
Console.Write(result - max_operations);
}
// Driver Code
public static void Main()
{
int N = 3;
int[] arr = { -1, 0, 2 };
findMinOperations(arr, N);
}
}
// This code is contributed by sanjoy_62.
<script>
// JavaScript code for the above approach
function findMinOperations(ar, n) {
// Initializing vector to avoid overflows
let arr = new Array(n + 5);
for (let i = 1; i <= n; i++) {
arr[i] = ar[i - 1];
}
let result = 0;
// Calculating minimum operations to be
// performed on initial array
for (let i = 2; i <= n; i++) {
result += Math.abs(arr[i] - arr[i - 1]);
}
// Way to change a value to make
// a1 equal to a2 or a(n)
// equal to a(n-1)
let max_operations
= Math.max(Math.abs(arr[1] - arr[2]),
Math.abs(arr[n] - arr[n - 1]));
for (let i = 2; i < n; i++) {
// For the rest of elements
// taking the max of
// operations already done +
// the ones performed here
max_operations
= Math.max(
max_operations,
Math.abs(arr[i] - arr[i - 1])
+ Math.abs(arr[i + 1] - arr[i])
- Math.abs(arr[i + 1] - arr[i - 1]));
}
// Print the final result
document.write((result - max_operations) + "</br>");
}
// Driver Code
let N = 3;
let arr = [-1, 0, 2];
findMinOperations(arr, N);
// This code is contributed by Potta Lokesh
</script>
Output
1
Time complexity: O(N)
Auxiliary Space: O(1)
