Minimum cost to process m tasks where switching costs
Last Updated :
21 Mar, 2025
Given an array of integers arr[] of size m, representing the different type of tasks that need to be processed. We need to find the minimum processing cost required to assign these tasks to n processor cores.
- Each core can process one task at a time.
- Assigning a task to a core incurs a cost of 1 if the task type is different from the task currently being processed by that core. If the core is already running the same type of task, the cost is 0.
- Initially, all cores are free, so the first task assigned to each core costs 1.
- A core continues processing its current task until a new task is assigned.
Example:
Input: n = 3, arr = [ 1, 2, 1, 3, 4, 1 ]
Output: 4
Explanation: Following are the states and cost required after each task:
There are 3 cores — A, B, and C.
- Assign task 1 to any core (A) → cost 1 unit.
- States: A - 1, B - None, C - None.
- Assign task 2 to another core (B) → cost 1 unit.
- States: A - 1, B - 2, C - None.
- Assign task 1 again to core A → cost 0 (same task type).
- States: A - 1, B - 2, C - None.
- Assign task 3 to the free core (C) → cost 1 unit.
- States: A - 1, B - 2, C - 3.
- All cores are now occupied. Assign task 4 to core B, replacing task 2 → cost 1 unit.
- States: A - 1, B - 4, C - 3.
- Finally, assign task 1 to core A again → cost 0 (same task type).
- States: A - 1, B - 4, C - 3.
Total cost: 1 + 1 + 0 + 1 + 1 + 0 = 4 units.
Input: n = 2, arr = [ 1, 2, 1, 3, 2, 1 ]
Output: 4
Explanation: Following are the states and cost required after each task:
There are 2 cores — A and B.
- Assign task 1 to any core (A) → cost 1 unit.
- Assign task 2 to the other core (B) → cost 1 unit.
- Assign task 1 again to core A → cost 0 (same task type).
- Assign task 3 to core A, replacing task 1 → cost 1 unit.
- Assign task 2 again to core B → cost 0 (same task type).
- Finally, assign task 1 to core A, replacing task 3 → cost 1 unit.
Total cost: 1 + 1 + 0 + 1 + 0 + 1 = 4 units.
Approach:
The idea is to manage which tasks are running on the cores at any given time. The main goal is to process each task while reducing unnecessary core switches, using a greedy approach to decide which task to stop when all cores are occupied.
The problem can be divided into two main cases:
- If the same type of task is already running in one of the cores:
- Assign the incoming task to that core — cost = 0.
- If the same type of task is not running in any core:
- Sub-case 1: If a core is free, assign the task to it — cost = 1.
- Sub-case 2: If all cores are occupied:
- Check if there is any running task that will never appear again — replace it.
- If all running tasks will reappear later, replace the task that will reoccur last — cost = 1.
Follow the below given steps to solve the problem:
- Create a 2D array
freq[][] of size m * m to store the future frequency of each task at every index. - Initialize
freqArr[] to track task frequencies from the last position, and fill freq[][] by iterating from the second last task to the first. - Use an
isRunning[] array to check which tasks are currently running on cores. - Track the total cost with a
cost variable and count occupied cores with count. - Iterate through each task:
- If the task is already running, continue — cost = 0.
- If a core is free, assign the task — cost = 1.
- If all cores are occupied:
- Check for a running task that will never reoccur — replace it and cost = 1.
- If all running tasks reoccur, find the one that appears farthest in the future — replace it and cost = 1.
- Update
isRunning[] after each task assignment.
Below is given the implementation:
C++
#include <bits/stdc++.h>
using namespace std;
// Function to find the minimum cost
// required to run m tasks on n cores
int minCost(int n, vector<int> &arr) {
int m = arr.size();
// to store the frequency of each
// task after the current position.
vector<vector<int>> freq(m);
// to store the frequency from last
vector<int> freqArr(m + 1, 0);
freq[m - 1] = freqArr;
// fill the freq array from last to first
for(int i = m - 2; i >= 0; i--) {
// add the frequency of the next task
freqArr[arr[i + 1]] += 1;
freq[i] = freqArr;
}
// to store if task is running
vector<int> isRunning(m + 1, 0);
// to store the total cost
int cost = 0;
// to store the count of occupied cores
int count = 0;
for(int i = 0; i < m; i++) {
// if task is already running,
// continue to the next task
if(isRunning[arr[i]]) {
continue;
}
// if task is not running, and
// there is a free core, use it
// and increment the cost
else if(count < n) {
isRunning[arr[i]] = 1;
count++;
cost++;
}
// if all cores are occupied
else {
// check if there is a task that is
// running and will not occur in future
bool flag = false;
for(int j = 1; j <= m; j++) {
if(isRunning[j] && freq[i][j] == 0) {
// stop the task and start the
// current task in that core
isRunning[j] = 0;
isRunning[arr[i]] = 1;
cost++;
flag = true;
break;
}
}
// if there is no such task is found
if(!flag) {
// find the farthest position where one of
// the currently running tasks will happen
int ind = m;
for(int j = m - 1; j > i; j--) {
if(isRunning[arr[j]]) {
ind = j;
break;
}
}
// stop that task and start the
// current task in that core
isRunning[arr[ind]] = 0;
isRunning[arr[i]] = 1;
cost++;
}
}
}
return cost;
}
int main() {
vector<int> arr = {1, 2, 1, 3, 4, 1};
int n = 3;
cout << minCost(n, arr);
return 0;
}
import java.util.*;
class GfG {
// Function to find the minimum cost
// required to run m tasks on n cores
static int minCost(int n, int[] arr) {
int m = arr.length;
// to store the frequency of each
// task after the current position.
int[][] freq = new int[m][m + 1];
// to store the frequency from last
int[] freqArr = new int[m + 1];
freq[m - 1] = freqArr.clone();
// fill the freq array from last to first
for (int i = m - 2; i >= 0; i--) {
// add the frequency of the next task
freqArr[arr[i + 1]] += 1;
freq[i] = freqArr.clone();
}
// to store if task is running
int[] isRunning = new int[m + 1];
// to store the total cost
int cost = 0;
// to store the count of occupied cores
int count = 0;
for (int i = 0; i < m; i++) {
// if task is already running,
// continue to the next task
if (isRunning[arr[i]] == 1) {
continue;
}
// if task is not running, and
// there is a free core, use it
// and increment the cost
else if (count < n) {
isRunning[arr[i]] = 1;
count++;
cost++;
}
// if all cores are occupied
else {
// check if there is a task that is
// running and will not occur in future
boolean flag = false;
for (int j = 1; j <= m; j++) {
if (isRunning[j] == 1 && freq[i][j] == 0) {
// stop the task and start the
// current task in that core
isRunning[j] = 0;
isRunning[arr[i]] = 1;
cost++;
flag = true;
break;
}
}
// if there is no such task is found
if (!flag) {
// find the farthest position where one of
// the currently running tasks will happen
int ind = m;
for (int j = m - 1; j > i; j--) {
if (isRunning[arr[j]] == 1) {
ind = j;
break;
}
}
// stop that task and start the
// current task in that core
isRunning[arr[ind]] = 0;
isRunning[arr[i]] = 1;
cost++;
}
}
}
return cost;
}
public static void main(String[] args) {
int[] arr = {1, 2, 1, 3, 4, 1};
int n = 3;
System.out.println(minCost(n, arr));
}
}
# Function to find the minimum cost
# required to run m tasks on n cores
def minCost(n, arr):
m = len(arr)
# to store the frequency of each
# task after the current position.
freq = [[0] * (m + 1) for _ in range(m)]
# to store the frequency from last
freqArr = [0] * (m + 1)
freq[m - 1] = freqArr[:]
# fill the freq array from last to first
for i in range(m - 2, -1, -1):
# add the frequency of the next task
freqArr[arr[i + 1]] += 1
freq[i] = freqArr[:]
# to store if task is running
isRunning = [0] * (m + 1)
# to store the total cost
cost = 0
# to store the count of occupied cores
count = 0
for i in range(m):
# if task is already running,
# continue to the next task
if isRunning[arr[i]]:
continue
# if task is not running, and
# there is a free core, use it
# and increment the cost
elif count < n:
isRunning[arr[i]] = 1
count += 1
cost += 1
# if all cores are occupied
else:
# check if there is a task that is
# running and will not occur in future
flag = False
for j in range(1, m + 1):
if isRunning[j] and freq[i][j] == 0:
# stop the task and start the
# current task in that core
isRunning[j] = 0
isRunning[arr[i]] = 1
cost += 1
flag = True
break
# if there is no such task is found
if not flag:
# find the farthest position where one of
# the currently running tasks will happen
ind = m
for j in range(m - 1, i, -1):
if isRunning[arr[j]]:
ind = j
break
# stop that task and start the
# current task in that core
isRunning[arr[ind]] = 0
isRunning[arr[i]] = 1
cost += 1
return cost
if __name__ == "__main__":
arr = [1, 2, 1, 3, 4, 1]
n = 3
print(minCost(n, arr))
using System;
using System.Collections.Generic;
class GfG {
// Function to find the minimum cost
// required to run m tasks on n cores
static int minCost(int n, List<int> arr) {
int m = arr.Count;
// to store the frequency of each
// task after the current position.
List<int[]> freq = new List<int[]>();
// to store the frequency from last
int[] freqArr = new int[m + 1];
freq.Add((int[])freqArr.Clone());
// fill the freq array from last to first
for (int i = m - 2; i >= 0; i--) {
freqArr[arr[i + 1]] += 1;
freq.Insert(0, (int[])freqArr.Clone());
}
// to store if task is running
int[] isRunning = new int[m + 1];
// to store the total cost
int cost = 0;
// to store the count of occupied cores
int count = 0;
for (int i = 0; i < m; i++) {
// if task is already running,
// continue to the next task
if (isRunning[arr[i]] == 1) {
continue;
}
// if task is not running, and
// there is a free core, use it
// and increment the cost
else if (count < n) {
isRunning[arr[i]] = 1;
count++;
cost++;
}
// if all cores are occupied
else {
// check if there is a task that is
// running and will not occur in future
bool flag = false;
for (int j = 1; j <= m; j++) {
if (isRunning[j] == 1 && freq[i][j] == 0) {
// stop the task and start the
// current task in that core
isRunning[j] = 0;
isRunning[arr[i]] = 1;
cost++;
flag = true;
break;
}
}
// if there is no such task is found
if (!flag) {
// find the farthest position where one of
// the currently running tasks will happen
int ind = m;
for (int j = m - 1; j > i; j--) {
if (isRunning[arr[j]] == 1) {
ind = j;
break;
}
}
// stop that task and start the
// current task in that core
isRunning[arr[ind]] = 0;
isRunning[arr[i]] = 1;
cost++;
}
}
}
return cost;
}
static void Main() {
List<int> arr = new List<int> {1, 2, 1, 3, 4, 1};
int n = 3;
Console.WriteLine(minCost(n, arr));
}
}
// Function to find the minimum cost
// required to run m tasks on n cores.
function minCost(n, arr) {
let m = arr.length;
// to store the frequency of each
// task after the current position.
let freq = new Array(m);
// to store the frequency from last
let freqArr = new Array(m + 1).fill(0);
freq[m - 1] = freqArr.slice();
// fill the freq array from last to first
for (let i = m - 2; i >= 0; i--) {
freqArr[arr[i + 1]] += 1;
freq[i] = freqArr.slice();
}
// to store if task is running
let isRunning = new Array(m + 1).fill(0);
// to store the total cost
let cost = 0;
// to store the count of occupied cores
let count = 0;
for (let i = 0; i < m; i++) {
// if task is already running,
// continue to the next task
if (isRunning[arr[i]] !== 0) {
continue;
}
// if task is not running, and
// there is a free core, use it
// and increment the cost
else if (count < n) {
isRunning[arr[i]] = 1;
count++;
cost++;
}
// if all cores are occupied
else {
// check if there is a task that is
// running and will not occur in future
let flag = false;
for (let j = 1; j <= m; j++) {
if (isRunning[j] !== 0 && freq[i][j] === 0) {
// stop the task and start the
// current task in that core
isRunning[j] = 0;
isRunning[arr[i]] = 1;
cost++;
flag = true;
break;
}
}
// if there is no such task is found
if (!flag) {
// find the farthest position where one of
// the currently running tasks will happen
let ind = m;
for (let j = m - 1; j > i; j--) {
if (isRunning[arr[j]] !== 0) {
ind = j;
break;
}
}
// stop that task and start the
// current task in that core
isRunning[arr[ind]] = 0;
isRunning[arr[i]] = 1;
cost++;
}
}
}
return cost;
}
// Driver Code
let arr = [1, 2, 1, 3, 4, 1];
let n = 3;
console.log(minCost(n, arr));
Time Complexity: O(m ^ 2), in the worst case, we are required to iterate through all the tasks for each of the m tasks, thus the overall time complexity will be O(m * m).
Space Complexity: O(m ^ 2), to create the 2d array freq[][] of order m * m.
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