Minimum cost to make parity of elements same by removing Subarray
Last Updated :
23 Jul, 2025
Given an arr[] of length N, the task is to make parity of arr[] the same by using the below-provided operation:
- Select a subarray containing elements of the same parity.
- Remove the subarray.
- The cost to remove that subarray is (absolute adjacent difference of all elements present in sub-array)*(length of subarray). For a sub-array of length 1, the cost will be the element present in that subarray.
Examples:
Input: N = 3, arr[] = {2, 4, 6}
Output: 0
Explanation: All the elements of given arr[] are even, Input arr[] has even parity already. Therefore, minimum cost is 0
Input: N = 4, arr[] = {22, 42, 64, 7}
Output: 7
Explanation: It will be optimal to remove sub-array {A4, . . . , A4} = {7}. Length of Sub-array is one, Therefore, minimum cost to remove this sub-array is = 7. After removing {7}, arr[] = {22, 42, 64}. it can be verified that now arr[] contains only even elements. Therefore, minimum cost to make parity of arr[] is 7.
Input: N = 7, arr[] = {2, 3, 1, 5, 4, 6, 4}
Output: 14
Explanation: It will be optimal to make arr[] of odd parity.
First sub-array = {2}, Cost = 2
Second sub-array = {4, 6, 4}, cost = ( |6-4|+|4-6| )*(3) = (2+2)*(3) = 12
Hence, Total minimum cost will be 2+12=14. arr[] after removing both sub-arrays = {3, 1, 5}
Approach: Implement the idea below to solve the problem:
The problem is based on Greedy approach for finding minimum cost. Find all the sub-arrays of even and odd parity, Then calculate minimum cost for both in two different variables. Print the minimum between both costs.
Follow the illustration below for a better understanding:
Illustr:
Consider array arr[] = {2, 3, 1, 5, 4, 6, 4}
Let us make the parity of given arr[] odd and even one by one.
- Even parity arr[]: For making arr[] of even parity, We have to remove all sub-arrays having odd parity along with their cost. There is only one odd subarray present in arr[] from index 1 to 3.
- First odd sub-array = {3, 1, 5}. Cost for removing this sub-array = ( |1-3|+|5-1| )*(3) = 6*3=18
- arr[] after removing the subarray is {2, 4, 6, 4}. It can be verified that now arr[] have even parity having costs as 18.
- Odd parity arr[]: For making arr[] of odd parity, We have to remove all sub-arrays having even parity along with their cost. There are two even subarrays present in arr[] from index 0 to 0 and 4 to 6 respectively.
- First even sub-array = {2}. Cost for removing this sub-array = 2
- Second even sub-array = {4, 6, 4}. Cost for removing this sub-array = ( |6-4|+|4-6| )*(3) = 4*3 = 12
- arr[] after removing sub-arrays {1} and {4, 6, 4} is {1, 3, 5}. It can be verified that now arr[] have odd parity having costs as 2+12=14.
Now, We have test arr[] for both parities. The minimum cost will be min(cost for making arr[] parity even, cost for making arr[] parity odd).
Minimum cost = min(14, 18) = 14.
Follow the steps mentioned below to implement the idea:
- Consider two variables (Say min_cost_even and min_cost_odd) for holding minimum cost to make parity of arr[] odd or even respectively.
- Find all subarrays having odd parity, Calculate the cost of each sub-array and add it into min_cost_even.
- Find all sub-arrays having Even parity, Calculate the cost of each sub-array and add it into min_cost_odd.
- Return the minimum value between both costs obtained in steps 2 and 3 i.e., min(min_cost_odd, min_cost_even).
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
// Function for finding minimum value
// from two input arguments
long mini(long a, long b) { return a <= b ? a : b; }
// Function for returning minimum cost
long minCost(int N, int arr[])
{
// Variable to hold minimum cost,
// to make arr[] parity even
long min_cost_even = 0;
// LeftEnd of arr[] initialized for finding
// subarrays containing same parity elements
int leftEnd = 0;
// Variable to hold minimum cost,
// to make arr[] parity odd
long min_cost_odd = 0;
// Algorithm to find Sub-arrays of Odd parity
// according to 0 based indexing
while (leftEnd < N) {
if (arr[leftEnd] % 2 == 0) {
leftEnd++;
}
else {
int rightEnd = leftEnd;
while (rightEnd < N - 1
&& arr[rightEnd + 1] % 2 != 0)
rightEnd = rightEnd + 1;
// Condition When sub-array has length 1
if (leftEnd == rightEnd) {
// Adding mincost to verify this
// sub-array
min_cost_even += arr[leftEnd];
}
// When length is greater than 1
else {
// Temporary variable to hold cost for
// this sub-array
long temp = 0;
// Loop for traversing on subarray
for (int i = leftEnd + 1; i <= rightEnd;
i++) {
// Initializing temp with absolute
// adjacent difference
temp += (abs(arr[i] - arr[i - 1]));
}
// Multiplying temp with subarray's
// length
temp *= (rightEnd - leftEnd + 1);
// Adding temp's value to min_cost_even
min_cost_even += temp;
}
// Incrementing leftEnd for finding next
// subarray of odd parity
leftEnd = rightEnd + 1;
}
}
// Set leftEnd to 0, So that we can traverse again
// on arr[] For finding even parity sub-arrays
leftEnd = 0;
// Algorithm to find Sub-arrays of Even parity
// according to 0 based indexing
while (leftEnd < N) {
if (arr[leftEnd] % 2 != 0) {
leftEnd++;
}
else {
int rightEnd = leftEnd;
while (rightEnd < N - 1
&& arr[rightEnd + 1] % 2 == 0)
rightEnd = rightEnd + 1;
// If sub-array is of length 1
if (leftEnd == rightEnd) {
// Adding cost to min_cost_odd variable
min_cost_odd += arr[leftEnd];
}
// When sub-array has length greater than 1
else {
// Temp variable to hold cost for this
// even parity sub-array
long temp = 0;
// Loop for traversing on sub-array
for (int i = leftEnd + 1; i <= rightEnd;
i++) {
// Adding absolute adjacent
// difference to temp
temp += (abs(arr[i] - arr[i - 1]));
}
// Multiplying temp with length of
// sub-array
temp *= (rightEnd - leftEnd + 1);
// Adding temp value to min_cost_odd
min_cost_odd += temp;
}
// Incrementing leftEnd for finding next
// Even parity sub-array
leftEnd = rightEnd + 1;
}
}
// Returning minimum cost
return mini(min_cost_odd, min_cost_even);
}
int main()
{
// Testcase1
int N = 7;
int arr[] = { 2, 3, 1, 5, 4, 6, 4 };
// Function call
cout << (minCost(N, arr)) << endl;
// Testcase2
int N2 = 5;
int arr2[] = { 1, 2, 3, 5, 4 };
// Function call
cout << (minCost(N2, arr2)) << endl;
return 0;
}
// This code is contributed by ksam24000
// Java code to implement the approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG {
// Driver Function
public static void main(String[] args)
{
// Testcase1
int N = 7;
int[] arr = { 2, 3, 1, 5, 4, 6, 4 };
// Function call
System.out.println(minCost(N, arr));
// Testcase2
int N2 = 5;
int[] arr2 = { 1, 2, 3, 5, 4 };
// Function call
System.out.println(minCost(N2, arr2));
}
// Function for returning minimum cost
static long minCost(int N, int[] arr)
{
// Variable to hold minimum cost,
// to make arr[] parity even
long min_cost_even = 0;
// LeftEnd of arr[] initialized for finding
// subarrays containing same parity elements
int leftEnd = 0;
// Variable to hold minimum cost,
// to make arr[] parity odd
long min_cost_odd = 0;
// Algorithm to find Sub-arrays of Odd parity
// according to 0 based indexing
while (leftEnd < N) {
if (arr[leftEnd] % 2 == 0) {
leftEnd++;
}
else {
int rightEnd = leftEnd;
while (rightEnd < N - 1
&& arr[rightEnd + 1] % 2 != 0)
rightEnd = rightEnd + 1;
// Condition When sub-array has length 1
if (leftEnd == rightEnd) {
// Adding mincost to verify this
// sub-array
min_cost_even += arr[leftEnd];
}
// When length is greater than 1
else {
// Temporary variable to hold cost for
// this sub-array
long temp = 0;
// Loop for traversing on subarray
for (int i = leftEnd + 1; i <= rightEnd;
i++) {
// Initializing temp with absolute
// adjacent difference
temp += (Math.abs(arr[i]
- arr[i - 1]));
}
// Multiplying temp with subarray's
// length
temp *= (rightEnd - leftEnd + 1);
// Adding temp's value to min_cost_even
min_cost_even += temp;
}
// Incrementing leftEnd for finding next
// subarray of odd parity
leftEnd = rightEnd + 1;
}
}
// Set leftEnd to 0, So that we can traverse again
// on arr[] For finding even parity sub-arrays
leftEnd = 0;
// Algorithm to find Sub-arrays of Even parity
// according to 0 based indexing
while (leftEnd < N) {
if (arr[leftEnd] % 2 != 0) {
leftEnd++;
}
else {
int rightEnd = leftEnd;
while (rightEnd < N - 1
&& arr[rightEnd + 1] % 2 == 0)
rightEnd = rightEnd + 1;
// If sub-array is of length 1
if (leftEnd == rightEnd) {
// Adding cost to min_cost_odd variable
min_cost_odd += arr[leftEnd];
}
// When sub-array has length greater than 1
else {
// Temp variable to hold cost for this
// even parity sub-array
long temp = 0;
// Loop for traversing on sub-array
for (int i = leftEnd + 1; i <= rightEnd;
i++) {
// Adding absolute adjacent
// difference to temp
temp += (Math.abs(arr[i]
- arr[i - 1]));
}
// Multiplying temp with length of
// sub-array
temp *= (rightEnd - leftEnd + 1);
// Adding temp value to min_cost_odd
min_cost_odd += temp;
}
// Incrementing leftEnd for finding next
// Even parity sub-array
leftEnd = rightEnd + 1;
}
}
// Returning minimum cost
return min(min_cost_odd, min_cost_even);
}
// Function for finding minimum value
// from two input arguments
static long min(long a, long b)
{
return a <= b ? a : b;
}
}
# Python code to implement the approach
# Function for finding minimum value
# from two input arguments
def Min(a, b):
return a if a <= b else b
# Function for returning minimum cost
def minCost(N, arr):
# Variable to hold minimum cost,
# to make arr[] parity even
min_cost_even = 0
# LeftEnd of arr[] initialized for finding
# subarrays containing same parity elements
leftEnd = 0
# Variable to hold minimum cost,
# to make arr[] parity odd
min_cost_odd = 0
# Algorithm to find Sub-arrays of Odd parity
# according to 0 based indexing
while(leftEnd < N):
if(arr[leftEnd] % 2 == 0):
leftEnd += 1
else:
rightEnd = leftEnd
while(rightEnd < N-1 and arr[rightEnd + 1] % 2 != 0):
rightEnd = rightEnd + 1
# Condition When sub-array has length 1
if(leftEnd == rightEnd):
# Adding mincost to verify this sub-array
min_cost_even = min_cost_even + arr[leftEnd]
# when length is greater than 1
else:
# Temporary variable to hold cost for this sub-array
temp = 0
# Loop for traversing on subarray
for i in range(leftEnd + 1, rightEnd+1):
# Initializing temp with absolute adjacent difference
temp = temp + abs(arr[i] - arr[i-1])
# Multiplying temp with subarray's length
temp = temp * (rightEnd - leftEnd + 1)
# Adding temp's value to min_cost_even
min_cost_even = min_cost_even + temp
# Incrementing leftEnd for finding next
# subarray of odd parity
leftEnd = rightEnd + 1
# Set leftEnd to 0, So that we can traverse again
# on arr[] For finding even parity sub-arrays
leftEnd = 0
# Algorithm to find Sub-arrays of Even parity
# according to 0 based indexing
while(leftEnd < N):
if(arr[leftEnd] % 2 != 0):
leftEnd += 1
else:
rightEnd = leftEnd
while(rightEnd < N-1 and arr[rightEnd + 1] % 2 == 0):
rightEnd = rightEnd + 1
# If sub-array is of length 1
if(leftEnd == rightEnd):
# Adding cost to min_cost_odd variable
min_cost_odd = min_cost_odd + arr[leftEnd]
# When sub-array has length greater than 1
else:
# Temp variable to hold cost for this even parity sub-array
temp = 0
# loop for traversing on sub-array
for i in range(leftEnd + 1, rightEnd+1):
# Adding absolute adjacent difference to temp
temp = temp + abs(arr[i] - arr[i-1])
# Multiplying temp with length of sub-array
temp = temp * (rightEnd - leftEnd + 1)
# Adding temp value to min_cost_odd
min_cost_odd = min_cost_odd + temp
# Incrementing leftEnd for finding next
# Even parity sub-array
leftEnd = rightEnd + 1
# Returning minimum cost
return Min(min_cost_odd, min_cost_even)
# Testcase1
N = 7
arr = [2, 3, 1, 5, 4, 6, 4]
# Function call
print(minCost(N, arr))
# Testcase2
N2 = 5
arr2 = [1, 2, 3, 5, 4]
# Function call
print(minCost(N2, arr2))
# This code is contributed by lokeshmvs21.
// C# code to implement the approach
using System;
public class GFG {
static public void Main()
{
// Testcase1
int N = 7;
int[] arr = { 2, 3, 1, 5, 4, 6, 4 };
// Function call
Console.WriteLine(minCost(N, arr));
// Testcase2
int N2 = 5;
int[] arr2 = { 1, 2, 3, 5, 4 };
// Function call
Console.WriteLine(minCost(N2, arr2));
}
// Function for returning minimum cost
static long minCost(int N, int[] arr)
{
// Variable to hold minimum cost,
// to make arr[] parity even
long min_cost_even = 0;
// LeftEnd of arr[] initialized for finding
// subarrays containing same parity elements
int leftEnd = 0;
// Variable to hold minimum cost,
// to make arr[] parity odd
long min_cost_odd = 0;
// Algorithm to find Sub-arrays of Odd parity
// according to 0 based indexing
while (leftEnd < N) {
if (arr[leftEnd] % 2 == 0) {
leftEnd++;
}
else {
int rightEnd = leftEnd;
while (rightEnd < N - 1
&& arr[rightEnd + 1] % 2 != 0)
rightEnd = rightEnd + 1;
// Condition When sub-array has length 1
if (leftEnd == rightEnd) {
// Adding mincost to verify this
// sub-array
min_cost_even += arr[leftEnd];
}
// When length is greater than 1
else {
// Temporary variable to hold cost for
// this sub-array
long temp = 0;
// Loop for traversing on subarray
for (int i = leftEnd + 1; i <= rightEnd;
i++) {
// Initializing temp with absolute
// adjacent difference
temp += (Math.Abs(arr[i]
- arr[i - 1]));
}
// Multiplying temp with subarray's
// length
temp *= (rightEnd - leftEnd + 1);
// Adding temp's value to min_cost_even
min_cost_even += temp;
}
// Incrementing leftEnd for finding next
// subarray of odd parity
leftEnd = rightEnd + 1;
}
}
// Set leftEnd to 0, So that we can traverse again
// on arr[] For finding even parity sub-arrays
leftEnd = 0;
// Algorithm to find Sub-arrays of Even parity
// according to 0 based indexing
while (leftEnd < N) {
if (arr[leftEnd] % 2 != 0) {
leftEnd++;
}
else {
int rightEnd = leftEnd;
while (rightEnd < N - 1
&& arr[rightEnd + 1] % 2 == 0)
rightEnd = rightEnd + 1;
// If sub-array is of length 1
if (leftEnd == rightEnd) {
// Adding cost to min_cost_odd variable
min_cost_odd += arr[leftEnd];
}
// When sub-array has length greater than 1
else {
// Temp variable to hold cost for this
// even parity sub-array
long temp = 0;
// Loop for traversing on sub-array
for (int i = leftEnd + 1; i <= rightEnd;
i++) {
// Adding absolute adjacent
// difference to temp
temp += (Math.Abs(arr[i]
- arr[i - 1]));
}
// Multiplying temp with length of
// sub-array
temp *= (rightEnd - leftEnd + 1);
// Adding temp value to min_cost_odd
min_cost_odd += temp;
}
// Incrementing leftEnd for finding next
// Even parity sub-array
leftEnd = rightEnd + 1;
}
}
// Returning minimum cost
return min(min_cost_odd, min_cost_even);
}
// Function for finding minimum value
// from two input arguments
static long min(long a, long b)
{
return a <= b ? a : b;
}
}
// This code is contributed by lokeshmvs21.
// JavaScript code for the above approach
// Function for returning minimum cost
function minCost(N, arr) {
// Variable to hold minimum cost,
// to make arr[] parity even
let min_cost_even = 0;
// LeftEnd of arr[] initialized for finding
// subarrays containing same parity elements
let leftEnd = 0;
// Variable to hold minimum cost,
// to make arr[] parity odd
let min_cost_odd = 0;
// Algorithm to find Sub-arrays of Odd parity
// according to 0 based indexing
while (leftEnd < N) {
if (arr[leftEnd] % 2 == 0) {
leftEnd++;
}
else {
let rightEnd = leftEnd;
while (rightEnd < N - 1
&& arr[rightEnd + 1] % 2 != 0)
rightEnd = rightEnd + 1;
// Condition When sub-array has length 1
if (leftEnd == rightEnd) {
// Adding mincost to verify this
// sub-array
min_cost_even += arr[leftEnd];
}
// When length is greater than 1
else {
// Temporary variable to hold cost for
// this sub-array
let temp = 0;
// Loop for traversing on subarray
for (let i = leftEnd + 1; i <= rightEnd;
i++) {
// Initializing temp with absolute
// adjacent difference
temp += (Math.abs(arr[i]
- arr[i - 1]));
}
// Multiplying temp with subarray's
// length
temp *= (rightEnd - leftEnd + 1);
// Adding temp's value to min_cost_even
min_cost_even += temp;
}
// Incrementing leftEnd for finding next
// subarray of odd parity
leftEnd = rightEnd + 1;
}
}
// Set leftEnd to 0, So that we can traverse again
// on arr[] For finding even parity sub-arrays
leftEnd = 0;
// Algorithm to find Sub-arrays of Even parity
// according to 0 based indexing
while (leftEnd < N) {
if (arr[leftEnd] % 2 != 0) {
leftEnd++;
}
else {
let rightEnd = leftEnd;
while (rightEnd < N - 1
&& arr[rightEnd + 1] % 2 == 0)
rightEnd = rightEnd + 1;
// If sub-array is of length 1
if (leftEnd == rightEnd) {
// Adding cost to min_cost_odd variable
min_cost_odd += arr[leftEnd];
}
// When sub-array has length greater than 1
else {
// Temp variable to hold cost for this
// even parity sub-array
let temp = 0;
// Loop for traversing on sub-array
for (let i = leftEnd + 1; i <= rightEnd;
i++) {
// Adding absolute adjacent
// difference to temp
temp += (Math.abs(arr[i]
- arr[i - 1]));
}
// Multiplying temp with length of
// sub-array
temp *= (rightEnd - leftEnd + 1);
// Adding temp value to min_cost_odd
min_cost_odd += temp;
}
// Incrementing leftEnd for finding next
// Even parity sub-array
leftEnd = rightEnd + 1;
}
}
// Returning minimum cost
return min(min_cost_odd, min_cost_even);
}
// Function for finding minimum value
// from two input arguments
function min(a, b) {
return a <= b ? a : b;
}
// Driver Function
// Testcase1
let N = 7;
let arr = [2, 3, 1, 5, 4, 6, 4];
// Function call
console.log(minCost(N, arr) + "<br>");
// Testcase2
let N2 = 5;
let arr2 = [1, 2, 3, 5, 4];
// Function call
console.log(minCost(N2, arr2));
// This code is contributed by Potta Lokesh
Time Complexity: O(N)
Auxiliary Space: No extra space is used.
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