 Open in App
Not now

# Minimum time required to complete all tasks without altering their order

• Difficulty Level : Medium
• Last Updated : 28 Dec, 2022

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 `   `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 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` `    ``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` `map``):`   `            ``# 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 ``+``=` `1`   `    ``# Print the result` `    ``print``(curr_time)`     `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:`   `    ``S ``=` `"ABACCA"` `    ``K ``=` `2` `    ``findMinimumTime(S, K)`   `  ``#  This code is contributed by ukasp.`

## 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

 ``

Output:

`9`

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

My Personal Notes arrow_drop_up
Related Articles