Given an array arr[] consisting of six integer digits only, the task is to return the maximum time in a 24-hour format that can be represented using the digits from the given array.
Note: The minimum time in 24-hour format is 00:00:00, and the maximum is 23:59:59. If a valid time cannot be formed, then print -1.
Examples:
Input: arr[] = {0, 2, 1, 9, 3, 2}
Output: 23:29:10
Explanation:
Maximum 24-Hour Format Time that can be formed using the digits of the array is 23:29:10Input: arr[] = {6, 2, 6, 7, 5, 6}
Output: -1
Approach: Follow the steps below to solve the problem:
- Create a Hashmap and store the frequency of digits in the given array.
- Iterate from maximum time 23:59:59 to the minimum time 00:00:00
- For every time, check if all the digits are in the Hashmap or not.
- Print the first time for which the above condition is found to be holding true. If no time is found to be satisfying the condition, print “-1”.
Below is the implementation of the above approach:
C++
// C++ Program of the// above approach#include <bits/stdc++.h>using namespace std;// Function to return the maximum// possible time in 24-Hours format// that can be represented by array elementsstring largestTimeFromDigits(vector<int>& A){ // Stores the frequency // of the array elements map<int, int> mp1, mp2; for (auto x : A) { mp1[x]++; } mp2 = mp1; // Maximum possible time int hr = 23, m = 59, s = 59; // Iterate to minimum possible time while (hr >= 0) { int h0 = hr / 10, h1 = hr % 10; int m0 = m / 10, m1 = m % 10; int s0 = s / 10, s1 = s % 10; int p = 0; vector<int> arr{ h0, h1, m0, m1, s0, s1 }; // Conditions to reduce the // the time iteratively for (auto& it : arr) { if (mp1[it] > 0) { mp1[it]--; } else { p = 1; } } // If all required digits // are present in the Map if (p == 0) { string s = ""; s = to_string(h0) + to_string(h1); s += ':' + to_string(m0) + to_string(m1); s += ':' + to_string(s0) + to_string(s1); return s; } // Retrieve Original Count mp1 = mp2; // If seconds is reduced to 0 if (s == 0) { s = 59; m--; } // If minutes is reduced to 0 else if (m < 0) { m = 59; hr--; } if (s > 0) { s--; } } return "-1";}// Driver Codeint main(){ vector<int> v = { 0, 2, 1, 9, 3, 2 }; cout << largestTimeFromDigits(v);} |
Java
// Java Program of the// above approachimport java.util.*;class GFG{// Function to return the maximum// possible time in 24-Hours format// that can be represented by array elementsstatic String largestTimeFromDigits(int []A){ // Stores the frequency // of the array elements HashMap<Integer, Integer> mp1 = new HashMap<Integer, Integer>(); HashMap<Integer, Integer> mp2 = new HashMap<Integer, Integer>(); for (int x : A) { if(mp1.containsKey(x)) mp1.put(x, mp1.get(x) + 1); else mp1.put(x, 1); } mp2 = (HashMap<Integer, Integer>) mp1.clone(); // Maximum possible time int hr = 23, m = 59, s = 59; // Iterate to minimum // possible time while (hr >= 0) { int h0 = hr / 10, h1 = hr % 10; int m0 = m / 10, m1 = m % 10; int s0 = s / 10, s1 = s % 10; int p = 0; int []arr = {h0, h1, m0, m1, s0, s1}; // Conditions to reduce the // the time iteratively for (int it : arr) { if (mp1.containsKey(it) && mp1.get(it) > 0) { mp1.put(it, mp1.get(it) - 1); } else { p = 1; } } // If all required digits // are present in the Map if (p == 0) { String st = ""; st = String.valueOf(h0) + String.valueOf(h1); st += ':' + String.valueOf(m0) + String.valueOf(m1); st += ':' + String.valueOf(s0) + String.valueOf(s1); return st; } // Retrieve Original Count mp1 = (HashMap<Integer, Integer>) mp2.clone(); // If seconds is reduced to 0 if (s == 0) { s = 59; m--; } // If minutes is reduced to 0 else if (m < 0) { m = 59; hr--; } if (s > 0) { s--; } } return "-1";}// Driver Codepublic static void main(String[] args){ int []v = {0, 2, 1, 9, 3, 2}; System.out.print(largestTimeFromDigits(v));}}// This code contributed by Princi Singh |
Python3
# Python 3 Program of the# above approach# Function to return the # maximum possible time in # 24-Hours format that can # be represented by array elementsdef largestTimeFromDigits(A): # Stores the frequency # of the array elements mp1 = {} mp2 = {} for x in A: mp1[x] = mp1.get(x, 0) + 1 mp2 = mp1.copy() # Maximum possible time hr = 23 m = 59 s = 59 # Iterate to minimum # possible time while(hr >= 0): h0 = hr//10 h1 = hr % 10 m0 = m//10 m1 = m%10 s0 = s//10 s1 = s%10 p = 0 arr = [h0, h1, m0, m1, s0, s1] # Conditions to reduce the # the time iteratively for it in arr: if (it in mp1 and mp1.get(it) > 0): mp1[it] = mp1.get(it) - 1 else: p = 1 # If all required digits # are present in the Map if (p == 0): s = "" s = (str(h0) + str(h1)) s += (':' + str(m0) + str(m1)) s += (':' + str(s0) + str(s1)) return s # Retrieve Original Count mp1 = mp2.copy() # If seconds is # reduced to 0 if (s == 0): s = 59 m -= 1 # If minutes is # reduced to 0 elif (m < 0): m = 59 hr -= 1 if (s > 0): s -= 1 return "-1"# Driver Codeif __name__ == '__main__': v = [0, 2, 1, 9, 3, 2] print(largestTimeFromDigits(v)) # This code is contributed by ipg2016107 |
C#
// C# Program of the// above approachusing System;using System.Collections.Generic;class GFG{// Function to return the maximum// possible time in 24-Hours format// that can be represented by array elementsstatic String largestTimeFromDigits(int []A){ // Stores the frequency // of the array elements Dictionary<int, int> mp1 = new Dictionary<int, int>(); Dictionary<int, int> mp2 = new Dictionary<int, int>(); foreach (int x in A) { if(mp1.ContainsKey(x)) mp1[x] = mp1[x] + 1; else mp1.Add(x, 1); } mp2 = new Dictionary<int, int>(mp1); // Maximum possible time int hr = 23, m = 59, s = 59; // Iterate to minimum // possible time while (hr >= 0) { int h0 = hr / 10, h1 = hr % 10; int m0 = m / 10, m1 = m % 10; int s0 = s / 10, s1 = s % 10; int p = 0; int []arr = {h0, h1, m0, m1, s0, s1}; // Conditions to reduce the // the time iteratively foreach (int it in arr) { if (mp1.ContainsKey(it) && mp1[it] > 0) { mp1[it]= mp1[it] - 1; } else { p = 1; } } // If all required digits // are present in the Map if (p == 0) { String st = ""; st = String.Join("", h0) + String.Join("", h1); st += ':' + String.Join("", m0) + String.Join("", m1); st += ':' + String.Join("", s0) + String.Join("", s1); return st; } // Retrieve Original Count mp1 = new Dictionary<int, int>(mp2); // If seconds is reduced to 0 if (s == 0) { s = 59; m--; } // If minutes is reduced to 0 else if (m < 0) { m = 59; hr--; } if (s > 0) { s--; } } return "-1";}// Driver Codepublic static void Main(String[] args){ int []v = {0, 2, 1, 9, 3, 2}; Console.Write(largestTimeFromDigits(v));}}// This code is contributed by shikhasingrajput |
Javascript
<script>class GFG{ // Function to return the maximum // possible time in 24-Hours format // that can be represented by array elements static largestTimeFromDigits(A) { // Stores the frequency // of the array elements var mp1 = new Map(); var mp2 = new Map(); for ( const x of A) { if (mp1.has(x)) { mp1.set(x,mp1.get(x) + 1); } else { mp1.set(x,1); }} mp2 = new Map(mp1); // Maximum possible time var hr = 23; var m = 59; var s = 59; // Iterate to minimum // possible time while (hr >= 0) { var h0 = parseInt(hr / 10); var h1 = hr % 10; var m0 = parseInt(m / 10); var m1 = m % 10; var s0 = parseInt(s / 10); var s1 = s % 10; var p = 0; var arr = [h0, h1, m0, m1, s0, s1]; // Conditions to reduce the // the time iteratively for ( const it of arr) { if (mp1.has(it) && mp1.get(it) > 0) { mp1.set(it,mp1.get(it) - 1); } else { p = 1; }} // If all required digits // are present in the Map if (p == 0) { var st = ""; st = new String(h0).toString() + new String(h1).toString(); st += ':' + new String(m0).toString() + new String(m1).toString(); st += ':' + new String(s0).toString() + new String(s1).toString(); return st; } // Retrieve Original Count mp1 = new Map(mp2); // If seconds is reduced to 0 if (s == 0) { s = 59; m--; } else if (m < 0) { m = 59; hr--; } if (s > 0) { s--; } } return "-1"; } // Driver Code static main(args) { var v = [0, 2, 1, 9, 3, 2]; document.write(GFG.largestTimeFromDigits(v)); }}GFG.main([]);</script> |
Output:
23:29:10
Time Complexity: O(60*60*60)
Auxiliary Space: O(1)
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