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 any order, given 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 = “AAABBB”, K = 2
Output: 8
Explanation:
Below are the order of task that is to be completed:
A ? B ? _ ? A ? B ? _ ? A ? B
Therefore, the total time required is 8 units.Input: S = “AAABBB”, K = 0
Output: 6
Approach: The given problem can be solved based on the following observations:
- To complete the tasks in minimum time, the idea is to find the maximum occurring character in the array and complete those tasks first.
- Let the maximum occurring character be X and its frequency be freq.
- Now, to complete tasks of type X first, there must be difference of K between them, which leaves a total number of (freq – 1) * K empty slots in between the tasks of X.
- Then, fill these empty slots by traversing the other tasks since the other tasks have frequency less than or equal to that of X.
Follow the steps below to solve the problem:
- Initialize a HashMap, say M, that stores the frequency of each character in the given string S.
- Initialize two variables, maxchar and maxfreq, that stores the maximum occurring character and its frequency respectively.
- Traverse the given string S and increment the frequency of S[i] in the map M and if it is greater than maxfreq, and update the value of maxchar to S[i] and the value of maxfreq to its frequency.
- Store the number of empty slots in a variable, say S, as (maxfreq – 1) * K.
- Traverse the HashMap, M and for every character other than maxchar, update the value of empty slots, S by subtracting minimum of (maxfreq – 1) and M[S[i]] from S.
- Store the required minimum time in a variable, ans as sum of N and S, if the value of S ? 0.
- After completing the above steps, print the value of ans as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the minimum time// required to complete the tasks if// the order of tasks can be changedvoid findMinimumTime(string& S, int N, int K){ // If there is no task, print 0 if (N == 0) { cout << 0; return; } // Store the maximum occurring // character and its frequency int maxfreq = INT_MIN; char maxchar; // Stores the frequency of each // character unordered_map<char, int> um; // Traverse the string S for (int i = 0; i < N; i++) { // Increment the frequency of // the current character um[S[i]]++; // Update maxfreq and maxchar if (um[S[i]] > maxfreq) { maxfreq = um[S[i]]; maxchar = S[i]; } } // Store the number of empty slots int emptySlots = (maxfreq - 1) * K; // Traverse the hashmap, um for (auto& it : um) { if (it.first == maxchar) continue; // Fill the empty slots emptySlots -= min(it.second, maxfreq - 1); } // Store the required result int ans = N + max(0, emptySlots); // Print the result cout << ans;}// Driver Codeint main(){ string S = "AAABBB"; int K = 2; int N = S.length(); findMinimumTime(S, N, K); return 0;} |
Java
// Java program for the above approachimport java.util.HashMap;import java.util.Map;class GFG{// Function to find the minimum time// required to complete the tasks if// the order of tasks can be changedstatic void findMinimumTime(String S, int N, int K){ // If there is no task, print 0 if (N == 0) { System.out.println(0); return; } // Store the maximum occurring // character and its frequency int maxfreq = Integer.MIN_VALUE; char maxchar = ' '; // Stores the frequency of each // character HashMap<Character, Integer> um = new HashMap<>(); // Traverse the string S for(int i = 0; i < N; i++) { // Increment the frequency of // the current character um.put(S.charAt(i), um.getOrDefault(S.charAt(i), 0) + 1); // Update maxfreq and maxchar if (um.get(S.charAt(i)) > maxfreq) { maxfreq = um.get(S.charAt(i)); maxchar = S.charAt(i); } } // Store the number of empty slots int emptySlots = (maxfreq - 1) * K; // Traverse the hashmap, um for(Map.Entry<Character, Integer> it : um.entrySet()) { if (it.getKey() == maxchar) continue; // Fill the empty slots emptySlots -= Math.min( it.getValue(), maxfreq - 1); } // Store the required result int ans = N + Math.max(0, emptySlots); // Print the result System.out.println(ans);}// Driver codepublic static void main(String[] args){ String S = "AAABBB"; int K = 2; int N = S.length(); findMinimumTime(S, N, K);}}// This code is contributed by abhinavjain194 |
Python3
# Python3 program for the above approachimport sysfrom collections import defaultdict# Function to find the minimum time# required to complete the tasks if# the order of tasks can be changeddef findMinimumTime(S, N, K): # If there is no task, print 0 if (N == 0): print(0) return # Store the maximum occurring # character and its frequency maxfreq = -sys.maxsize # Stores the frequency of each # character um = defaultdict(int) # Traverse the string S for i in range(N): # Increment the frequency of # the current character um[S[i]] += 1 # Update maxfreq and maxchar if (um[S[i]] > maxfreq): maxfreq = um[S[i]] maxchar = S[i] # Store the number of empty slots emptySlots = (maxfreq - 1) * K # Traverse the hashmap, um for it in um: if (it == maxchar): continue # Fill the empty slots emptySlots -= min(um[it], maxfreq - 1) # Store the required result ans = N + max(0, emptySlots) # Print the result print(ans)# Driver Codeif __name__ == "__main__": S = "AAABBB" K = 2 N = len(S) findMinimumTime(S, N, K) # This code is contributed by ukasp |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{// Function to find the minimum time// required to complete the tasks if// the order of tasks can be changedstatic void findMinimumTime(string S, int N, int K){ // If there is no task, print 0 if (N == 0) { Console.Write(0); return; } // Store the maximum occurring // character and its frequency int maxfreq = Int32.MinValue; char maxchar = ' '; // Stores the frequency of each // character Dictionary<char, int> um = new Dictionary<char, int>(); // Traverse the string S for(int i = 0; i < N; i++) { // Increment the frequency of // the current character if (um.ContainsKey(S[i])) um[S[i]]++; else um.Add(S[i], 1); // Update maxfreq and maxchar if (um.ContainsKey(S[i]) && um[S[i]] > maxfreq) { maxfreq = um[S[i]]; maxchar = S[i]; } } // Store the number of empty slots int emptySlots = (maxfreq - 1) * K; // Traverse the hashmap, um foreach(KeyValuePair<char, int> kvp in um) { if (kvp.Key == maxchar) continue; // Fill the empty slots emptySlots -= Math.Min(kvp.Value, maxfreq - 1); } // Store the required result int ans = N + Math.Max(0, emptySlots); // Print the result Console.Write(ans);}// Driver Codepublic static void Main(){ string S = "AAABBB"; int K = 2; int N = S.Length; findMinimumTime(S, N, K);}}// This code is contributed by bgangwar59 |
Javascript
<script>// Javascript program for the above approach// Function to find the minimum time// required to complete the tasks if// the order of tasks can be changedfunction findMinimumTime(s, N, K){ // If there is no task, print 0 if (N == 0) { document.write(0); return; } // Store the maximum occurring // character and its frequency let maxfreq = Number.MIN_SAFE_INTEGER; let maxchar; // Stores the frequency of each // character let um = new Map(); // Traverse the string S for(let i = 0; i < N; i++) { // Increment the frequency of // the current character if (um.has(s[i])) { um.set(s[i], um.get(s[i]) + 1) } else { um.set(s[i], 1) } // Update maxfreq and maxchar if (um.get(S[i]) > maxfreq) { maxfreq = um.get(S[i]); maxchar = S[i]; } } // Store the number of empty slots let emptySlots = (maxfreq - 1) * K; // Traverse the hashmap, um for(let it of um) { if (it[0] == maxchar) continue; // Fill the empty slots emptySlots -= Math.min( it[1], maxfreq - 1); } // Store the required result let ans = N + Math.max(0, emptySlots); // Print the result document.write(ans);}// Driver Codelet S = "AAABBB";let K = 2;let N = S.length;findMinimumTime(S, N, K);// This code is contributed by _saurabh_jaiswal.</script> |
Output:
8
Time Complexity: O(N)
Auxiliary Space: O(256)
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