Maximum possible time that can be formed from four digits in C++



In this tutorial, we will be discussing a program to find maximum possible time that can be formed from four digits.

For this we will be provided with an array consisting 4 digits. Our task is to find the maximum time (24 hour format) that can formed using those four digits.

Example

 Live Demo

#include <bits/stdc++.h> using namespace std; //returning updated frequency map map<int, int> getFrequencyMap(int arr[], int n) {    map<int, int> hashMap;    for (int i = 0; i < n; i++) {       hashMap[arr[i]]++;    }    return hashMap; } //checking if the digit is present in frequency map bool hasDigit(map<int, int>* hashMap, int digit) {    if ((*hashMap)[digit]) {       (*hashMap)[digit]--;       return true;    }    return false; } //returning maximum time in 24 hour format string getMaxtime_value(int arr[], int n) {    map<int, int> hashMap = getFrequencyMap(arr, n);    int i;    bool flag;    string time_value = "";    flag = false;    for (i = 2; i >= 0; i--) {       if (hasDigit(&hashMap, i)) {          flag = true;          time_value += (char)i + 48;          break;       }    }    if (!flag)       return "-1";    flag = false;    if (time_value[0] == '2') {       for (i = 3; i >= 0; i--) {          if (hasDigit(&hashMap, i)) {             flag = true;             time_value += (char)i + 48;             break;          }       }    }    else {       for (i = 9; i >= 0; i--) {          if (hasDigit(&hashMap, i)) {             flag = true;             time_value += (char)i + 48;             break;          }       }    }    if (!flag)       return "-1";    time_value += ":";    flag = false;    for (i = 5; i >= 0; i--) {       if (hasDigit(&hashMap, i)) {          flag = true;          time_value += (char)i + 48;          break;       }    }    if (!flag)       return "-1";    flag = false;    for (i = 9; i >= 0; i--) {       if (hasDigit(&hashMap, i)) {          flag = true;          time_value += (char)i + 48;          break;       }    }    if (!flag)       return "-1";    return time_value; } int main() {    int arr[] = { 0, 0, 0, 9 };    int n = sizeof(arr) / sizeof(int);    cout << (getMaxtime_value(arr, n));    return 0; }

Output

09:00
Updated on: 2020-09-09T12:44:22+05:30

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