Maximum possible time that can be formed from four digits

02 Jul, 2022
Given an array arr[] having 4 integer digits only. The task is to return the maximum 24 hour time that can be formed using the digits from the array. Note that the minimum time in 24 hour format is 00:00, and the maximum is 23:59. If a valid time cannot be formed then return -1.Examples:  
Input: arr[] = {1, 2, 3, 4} Output: 23:41Input: arr[] = {5, 5, 6, 6} Output: -1  
 
Approach: Create a HashMap and store the frequency of each digit in the map which can be used to know how many of such digits are available. Now, in order to generate a valid time following conditions must be satisfied:  
First digit of hours must be from the range [0, 2]. Start checking in decreasing order in order to maximize the time i.e. from 2 to 0. Once the digit is chosen, decrement its occurrence in the map by 1.
Second digit of hours must be from the range [0, 3] if first digit was chosen as 2 else [0, 9]. Update the HashMap accordingly after choosing the digit.
First digit of minutes must be from the range [0, 5] and second digit of minutes must be from the range [0, 9].
If any of the above condition fails i.e. no digit could be chosen at any point then print -1 else print the time.Below is the implementation of the above approach:  
C++
Java
Python3
C#
Javascript
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std; // Function to return the updated frequency map// for the array passed as argumentmap<int, int> getFrequencyMap(int arr[], int n){    map<int, int> hashMap;    for (int i = 0; i < n; i++) {         hashMap[arr[i]]++;    }    return hashMap;} // Function that returns true if the passed digit is present// in the map after decrementing it's frequency by 1bool hasDigit(map<int, int>* hashMap, int digit){     // If map contains the digit    if ((*hashMap)[digit]) {         // Decrement the frequency of the digit by 1        (*hashMap)[digit]--;         // True here indicates that the digit was found in the map        return true;    }     // Digit not found    return false;} // Function to return the maximum possible time_value in 24-Hours formatstring getMaxtime_value(int arr[], int n){    map<int, int> hashMap = getFrequencyMap(arr, n);    int i;    bool flag;    string time_value = "";     flag = false;     // First digit of hours can be from the range [0, 2]    for (i = 2; i >= 0; i--) {        if (hasDigit(&hashMap, i)) {            flag = true;            time_value += (char)i + 48;            break;        }    }     // If no valid digit found    if (!flag)        return "-1";     flag = false;     // If first digit of hours was chosen as 2 then    // the second digit of hours can be    // from the range [0, 3]    if (time_value[0] == '2') {        for (i = 3; i >= 0; i--) {            if (hasDigit(&hashMap, i)) {                flag = true;                time_value += (char)i + 48;                break;            }        }    }     // Else it can be from the range [0, 9]    else {        for (i = 9; i >= 0; i--) {            if (hasDigit(&hashMap, i)) {                flag = true;                time_value += (char)i + 48;                break;            }        }    }    if (!flag)        return "-1";     // Hours and minutes separator    time_value += ":";     flag = false;     // First digit of minutes can be from the range [0, 5]    for (i = 5; i >= 0; i--) {        if (hasDigit(&hashMap, i)) {            flag = true;            time_value += (char)i + 48;            break;        }    }    if (!flag)        return "-1";     flag = false;     // Second digit of minutes can be from the range [0, 9]    for (i = 9; i >= 0; i--) {        if (hasDigit(&hashMap, i)) {            flag = true;            time_value += (char)i + 48;            break;        }    }    if (!flag)        return "-1";     // Return the maximum possible time_value    return time_value;} // Driver codeint main(){    int arr[] = { 0, 0, 0, 9 };    int n = sizeof(arr) / sizeof(int);    cout << (getMaxtime_value(arr, n));    return 0;}// contributed by Arnab Kundu
// Java implementation of the approach import java.util.*; public class GFG {     // Function to return the updated frequency map    // for the array passed as argument    static HashMap<Integer, Integer> getFrequencyMap(int arr[])    {        HashMap<Integer, Integer> hashMap = new HashMap<>();        for (int i = 0; i < arr.length; i++) {            if (hashMap.containsKey(arr[i])) {                hashMap.put(arr[i], hashMap.get(arr[i]) + 1);            }            else {                hashMap.put(arr[i], 1);            }        }        return hashMap;    }     // Function that returns true if the passed digit is present    // in the map after decrementing it's frequency by 1    static boolean hasDigit(HashMap<Integer, Integer> hashMap, int digit)    {         // If map contains the digit        if (hashMap.containsKey(digit) && hashMap.get(digit) > 0) {             // Decrement the frequency of the digit by 1            hashMap.put(digit, hashMap.get(digit) - 1);             // True here indicates that the digit was found in the map            return true;        }         // Digit not found        return false;    }     // Function to return the maximum possible time in 24-Hours format    static String getMaxTime(int arr[])    {        HashMap<Integer, Integer> hashMap = getFrequencyMap(arr);        int i;        boolean flag;        String time = "";         flag = false;         // First digit of hours can be from the range [0, 2]        for (i = 2; i >= 0; i--) {            if (hasDigit(hashMap, i)) {                flag = true;                time += i;                break;            }        }         // If no valid digit found        if (!flag) {            return "-1";        }         flag = false;         // If first digit of hours was chosen as 2 then        // the second digit of hours can be        // from the range [0, 3]        if (time.charAt(0) == '2') {            for (i = 3; i >= 0; i--) {                if (hasDigit(hashMap, i)) {                    flag = true;                    time += i;                    break;                }            }        }         // Else it can be from the range [0, 9]        else {            for (i = 9; i >= 0; i--) {                if (hasDigit(hashMap, i)) {                    flag = true;                    time += i;                    break;                }            }        }        if (!flag) {            return "-1";        }         // Hours and minutes separator        time += ":";         flag = false;         // First digit of minutes can be from the range [0, 5]        for (i = 5; i >= 0; i--) {            if (hasDigit(hashMap, i)) {                flag = true;                time += i;                break;            }        }        if (!flag) {            return "-1";        }         flag = false;         // Second digit of minutes can be from the range [0, 9]        for (i = 9; i >= 0; i--) {            if (hasDigit(hashMap, i)) {                flag = true;                time += i;                break;            }        }        if (!flag) {            return "-1";        }         // Return the maximum possible time        return time;    }     // Driver code    public static void main(String[] args)    {        int arr[] = { 0, 0, 0, 9 };        System.out.println(getMaxTime(arr));    }}
# Python3 implementation of the approachfrom collections import defaultdict # Function to return the updated frequency# map for the array passed as argumentdef getFrequencyMap(arr, n):         hashMap = defaultdict(lambda:0)    for i in range(n):        hashMap[arr[i]] += 1             return hashMap     # Function that returns true if the passed# digit is present in the map after# decrementing it's frequency by 1def hasDigit(hashMap, digit):         # If map contains the digit    if hashMap[digit] > 0:             # Decrement the frequency of        # the digit by 1        hashMap[digit] -= 1             # True here indicates that the        # digit was found in the map        return True         # Digit not found    return False     # Function to return the maximum possible# time_value in 24-Hours formatdef getMaxtime_value(arr, n):         hashMap = getFrequencyMap(arr, n)    flag = False    time_value = ""         # First digit of hours can be    # from the range [0, 2]    for i in range(2, -1, -1):        if hasDigit(hashMap, i) == True:            flag = True            time_value += str(i)            break         # If no valid digit found    if not flag:        return "-1"         flag = False         # If first digit of hours was chosen as 2 then    # the second digit of hours can be    # from the range [0, 3]    if(time_value[0] == '2'):        for i in range(3, -1, -1):            if hasDigit(hashMap, i) == True:                flag = True                time_value += str(i)                break         # Else it can be from the range [0, 9]    else:        for i in range(9, -1, -1):            if hasDigit(hashMap, i) == True:                flag = True                time_value += str(i)                break                 if not flag:        return "-1"         # Hours and minutes separator    time_value += ":"         flag = False         # First digit of minutes can be    # from the range [0, 5]    for i in range(5, -1, -1):        if hasDigit(hashMap, i) == True:            flag = True            time_value += str(i)            break             if not flag:        return "-1"         flag = False         # Second digit of minutes can be    # from the range [0, 9]    for i in range(9, -1, -1):        if hasDigit(hashMap, i) == True:            flag = True            time_value += str(i)            break             if not flag:        return "-1"         # Return the maximum possible    # time_value    return time_value     # Driver codeif __name__ == "__main__":         arr = [0, 0, 0, 9]    n = len(arr)    print(getMaxtime_value(arr, n))     # This code is contributed by# Rituraj Jain
// C# implementation of the approachusing System;using System.Collections.Generic; class GFG{     // Function to return the updated frequency map    // for the array passed as argument    static Dictionary<int, int> getFrequencyMap(int []arr)    {        Dictionary<int, int> hashMap = new Dictionary<int, int>();        for (int i = 0; i < arr.Length; i++)        {            if (hashMap.ContainsKey(arr[i]))            {                hashMap[arr[i]] = hashMap[arr[i]] + 1;            }            else            {                hashMap.Add(arr[i], 1);            }        }        return hashMap;    }     // Function that returns true if the passed digit is present    // in the map after decrementing it's frequency by 1    static bool hasDigit(Dictionary<int, int> hashMap, int digit)    {         // If map contains the digit        if (hashMap.ContainsKey(digit) && hashMap[digit] > 0)        {             // Decrement the frequency of the digit by 1            hashMap[digit] = hashMap[digit] - 1;             // True here indicates that the            // digit was found in the map            return true;        }         // Digit not found        return false;    }     // Function to return the maximum    // possible time in 24-Hours format    static String getMaxTime(int []arr)    {        Dictionary<int, int> hashMap = getFrequencyMap(arr);        int i;        bool flag;        String time = "";         flag = false;         // First digit of hours can be from the range [0, 2]        for (i = 2; i >= 0; i--)        {            if (hasDigit(hashMap, i))            {                flag = true;                time += i;                break;            }        }         // If no valid digit found        if (!flag)        {            return "-1";        }         flag = false;         // If first digit of hours was chosen as 2 then        // the second digit of hours can be        // from the range [0, 3]        if (time[0] == '2')        {            for (i = 3; i >= 0; i--)            {                if (hasDigit(hashMap, i))                {                    flag = true;                    time += i;                    break;                }            }        }         // Else it can be from the range [0, 9]        else        {            for (i = 9; i >= 0; i--)            {                if (hasDigit(hashMap, i))                {                    flag = true;                    time += i;                    break;                }            }        }        if (!flag)        {            return "-1";        }         // Hours and minutes separator        time += ":";         flag = false;         // First digit of minutes can be from the range [0, 5]        for (i = 5; i >= 0; i--)        {            if (hasDigit(hashMap, i))            {                flag = true;                time += i;                break;            }        }        if (!flag)        {            return "-1";        }         flag = false;         // Second digit of minutes can be from the range [0, 9]        for (i = 9; i >= 0; i--)        {            if (hasDigit(hashMap, i))            {                flag = true;                time += i;                break;            }        }        if (!flag)        {            return "-1";        }         // Return the maximum possible time        return time;    }     // Driver code    public static void Main(String[] args)    {        int []arr = { 0, 0, 0, 9 };        Console.WriteLine(getMaxTime(arr));    }} // This code is contributed by Rajput-Ji
<script>       // JavaScript implementation of the approach             // Function to return the updated frequency map      // for the array passed as argument      function getFrequencyMap(arr) {        var hashMap = {};        for (var i = 0; i < arr.length; i++) {          if (hashMap.hasOwnProperty(arr[i])) {            hashMap[arr[i]] = hashMap[arr[i]] + 1;          } else {            hashMap[arr[i]] = 1;          }        }        return hashMap;      }       // Function that returns true if the passed digit is present      // in the map after decrementing it's frequency by 1      function hasDigit(hashMap, digit) {        // If map contains the digit        if (hashMap.hasOwnProperty(digit) && hashMap[digit] > 0) {          // Decrement the frequency of the digit by 1          hashMap[digit] = hashMap[digit] - 1;           // True here indicates that the          // digit was found in the map          return true;        }         // Digit not found        return false;      }       // Function to return the maximum      // possible time in 24-Hours format      function getMaxTime(arr) {        var hashMap = getFrequencyMap(arr);        var i;        var flag;        var time = "";         flag = false;         // First digit of hours can be from the range [0, 2]        for (i = 2; i >= 0; i--) {          if (hasDigit(hashMap, i)) {            flag = true;            time += i;            break;          }        }         // If no valid digit found        if (!flag) {          return "-1";        }         flag = false;         // If first digit of hours was chosen as 2 then        // the second digit of hours can be        // from the range [0, 3]        if (time[0] === "2") {          for (i = 3; i >= 0; i--) {            if (hasDigit(hashMap, i)) {              flag = true;              time += i;              break;            }          }        }         // Else it can be from the range [0, 9]        else {          for (i = 9; i >= 0; i--) {            if (hasDigit(hashMap, i)) {              flag = true;              time += i;              break;            }          }        }        if (!flag) {          return "-1";        }         // Hours and minutes separator        time += ":";         flag = false;         // First digit of minutes can be from the range [0, 5]        for (i = 5; i >= 0; i--) {          if (hasDigit(hashMap, i)) {            flag = true;            time += i;            break;          }        }        if (!flag) {          return "-1";        }         flag = false;         // Second digit of minutes can be from the range [0, 9]        for (i = 9; i >= 0; i--) {          if (hasDigit(hashMap, i)) {            flag = true;            time += i;            break;          }        }        if (!flag) {          return "-1";        }         // Return the maximum possible time        return time;      }       // Driver code      var arr = [0, 0, 0, 9];      document.write(getMaxTime(arr));        </script>
09:00
 
Time Complexity: O(n), since the loop runs from 0 to (n – 1). 
Auxiliary Space: O(n), since n extra space has been taken.
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