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 ordervoid 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 Codeint main(){ string S = "ABACCA"; int K = 2; findMinimumTime(S, K); return 0;}// This code is contributed by Kingash. |
Java
// Java program for the above approachimport 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 orderdef 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 Codeif __name__ == "__main__": S = "ABACCA" K = 2 findMinimumTime(S, K) # This code is contributed by ukasp. |
C#
// C# implementation of// the above approachusing System;using System.Collections.Generic;class GFG{// Function to find the minimum// time required to complete// tasks without changing their orderstatic 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 Codepublic 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 orderfunction 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 Codevar 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)
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