Minimum time required to complete all tasks without altering their order

Given a string S consisting of N characters (representing the tasks to perform) and a positive integer K, the task is to find the minimum time required to complete all the given tasks in the given order such that each task takes one unit of time and each task of the same type must be performed at an interval of K units.

Examples:

Input: S = “ABACCA”, K = 2
Output: 9
Explanation:
Below are the order of task that is to be completed:
A ? B ? _ ? A ? C ? _ ? _ ? C ? A.
Therefore, the total time required is 9 units.

Input: S = “AAAA”, K = 3
Output: 13  

Approach: The given problem can be solved by Hashing by keeping track of the last instant of each task and find the overall minimum time accordingly. Follow the steps below to solve the problem:

• Initialize a hashmap, say M, to keep track of the last time instant of each task.
• Initialize a variable, say ans as 0, to store the resultant minimum time.
• Traverse the given string S and perform the following steps:
  • If the character S[i] is present in M, then store the last instant of S[i] in a variable, say t.
  • If the value of (ans – t) is at most K, then add the value of K – (ans – t) + 1 to variable ans.
  • Update the last time instant of the character S[i] in M to ans, and increment the value of ans by 1.
• After completing the above steps, print the value of ans as the result.

Below is the implementation of the above approach:

C++

// C++ implementation of
// the above approach
#include <bits/stdc++.h>

using namespace std;

// Function to find the minimum
// time required to complete
// tasks without changing their order
void findMinimumTime(string tasks, int K)
{
    // Keeps track of the last
    // time instant of each task
    unordered_map<char, int> map;

    // Stores the required result
    int curr_time = 0;

    // Traverse the given string
    for (char c : tasks) {

        // Check last time instant of
        // task, if it exists before
        if (map.find(c) != map.end()) {

            // Increment the time
            // if task is within
            // the K units of time
            if (curr_time - map <= K) {

                curr_time += K - (curr_time - map) + 1;
            }
        }

        // Update the time of the
        // current task in the map
        map = curr_time;

        // Increment the time by 1
        curr_time++;
    }

    // Print the result
    cout << curr_time;
}

// Driver Code
int main()
{
    string S = "ABACCA";
    int K = 2;
    findMinimumTime(S, K);

    return 0;
}

// This code is contributed by Kingash.

Java

// Java program for the above approach

import java.util.*;

class GFG {

    // Function to find the minimum
    // time required to complete
    // tasks without changing their order
    static void findMinimumTime(
        String tasks, int K)
    {
        // Keeps track of the last
        // time instant of each task
        Map<Character, Integer> map
            = new HashMap<>();

        // Stores the required result
        int curr_time = 0;

        // Traverse the given string
        for (char c : tasks.toCharArray()) {

            // Check last time instant of
            // task, if it exists before
            if (map.containsKey(c)) {

                // Increment the time
                // if task is within
                // the K units of time
                if (curr_time
                        - map.get(c)
                    <= K) {

                    curr_time += K
                                 - (curr_time
                                    - map.get(c))
                                 + 1;
                }
            }

            // Update the time of the
            // current task in the map
            map.put(c, curr_time);

            // Increment the time by 1
            curr_time++;
        }

        // Print the result
        System.out.println(curr_time);
    }

    // Driver Code
    public static void main(String[] args)
    {
        String S = "ABACCA";
        int K = 2;
        findMinimumTime(S, K);
    }
}

Python3

# Python 3 implementation of
# the above approach

# Function to find the minimum
# time required to complete
# tasks without changing their order
def findMinimumTime(tasks, K):
    # Keeps track of the last
    # time instant of each task
    last_time_map = {}

    # Stores the required result
    curr_time = 0

    # Traverse the given string
    for c in tasks:
        # Check last time instant of
        # task, if it exists before
        if c in last_time_map:
            # Increment the time
            # if task is within
            # the K units of time
            if curr_time - last_time_map <= K:
                curr_time += K - (curr_time - last_time_map) + 1

        # Update the time of the
        # current task in the map
        last_time_map = curr_time

        # Increment the time by 1
        curr_time += 1

    # Print the result
    print(curr_time)

# Driver Code
if __name__ == "__main__":
    S = "ABACCA"
    K = 2
    findMinimumTime(S, K)

C#

// C# implementation of
// the above approach
using System;
using System.Collections.Generic;

class GFG{

// Function to find the minimum
// time required to complete
// tasks without changing their order
static void findMinimumTime(String tasks, int K)
{

    // Keeps track of the last
    // time instant of each task
    Dictionary<char,
               int> map = new Dictionary<char,
                                         int>();

    // Stores the required result
    int curr_time = 0;

    // Traverse the given string
    foreach (char c in tasks.ToCharArray())
    {

        // Check last time instant of
        // task, if it exists before
        if (map.ContainsKey(c))
        {

            // Increment the time
            // if task is within
            // the K units of time
            if (curr_time - map <= K)
            {
                curr_time += K - (curr_time - map) + 1;
            }

        }

        // Update the time of the
        // current task in the map
        if (!map.ContainsKey(c))
            map.Add(c, curr_time);
        else
            map = curr_time;

        // Increment the time by 1
        curr_time++;
    }

    // Print the result
    Console.WriteLine(curr_time);
}

// Driver Code
public static void Main(String[] args)
{
    String S = "ABACCA";
    int K = 2;

    findMinimumTime(S, K);
}
}

// This code is contributed by shikhasingrajput

Javascript

<script>

// Javascript implementation of
// the above approach


// Function to find the minimum
// time required to complete
// tasks without changing their order
function findMinimumTime(tasks, K)
{
    // Keeps track of the last
    // time instant of each task
    var map = new Map();

    // Stores the required result
    var curr_time = 0;

    // Traverse the given string
    tasks.split('').forEach(c => {

        // Check last time instant of
        // task, if it exists before
        if (map.has(c)) {

            // Increment the time
            // if task is within
            // the K units of time
            if (curr_time - map.get(c) <= K) {

                curr_time += K - (curr_time - map.get(c)) + 1;
            }
        }

        // Update the time of the
        // current task in the map
        map.set(c, curr_time);

        // Increment the time by 1
        curr_time++;
    });

    // Print the result
    document.write( curr_time);
}

// Driver Code
var S = "ABACCA";
var K = 2;
findMinimumTime(S, K);

// This code is contributed by rutvik_56.
</script>

Output

9

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