Maximum time such that absolute difference between hour and minute lies in given range
Given a 24-hour time value, where on some numbers are question marks( ‘?’ ), and two integers L and R. The question mark can be replaced with any number. The task is to find the maximum time such that the absolute difference between hour value and minute value lies in the interval [L, R]. If no possible value of time exists, then print -1.
Examples:
Input : time = “2?:03”, L = 5, R = 18
Output : 21:03
Explanation : Since, the difference is 21 – 3 = 18 and the time value 21:03 is the largest possible value whose difference lies in the range [5, 18]Input : time = “??:??”, L = 60, R = 69
Output : -1
Explanation : Since maximum possible difference between hour value and minute value is 59. So, no time value is possible.
Approach :
We will run two nested loops, one to represent ‘hour’ value from 23 to 0 and another to represent ‘minute’ value from 59 to 0. We keep decreasing ‘hour’ and ‘minute’ value until we get the desired time value.
- Since the maximum time value is desired, so we decrease the ‘hour’ value from 23 to 0 and similarly ‘minute’ value from 59 to 0.
- After decreasing the ‘hour’ and ‘minute’ value, we need to check if the ‘hour’ and ‘minute’ value is valid or not.
- Change in value of ‘hour’ and ‘minute’ is allowed only at those positions which have ‘?’. Changes at other positions will make time value invalid. To ensure this we will call function isValid().
- Keep on decreasing ‘hour’ and ‘minute’ value until a valid time is found whose difference lies in the range [ L, R ].
- If no valid time is found, then print “-1”.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function checks whether // given time is correct bool isValid( char a1, char a2, string str, int flag) { char v1, v2; // To check minute value of time if (flag == 0) { v1 = str[4]; v2 = str[3]; } else { // To check hour value of time v1 = str[1]; v2 = str[0]; } // Changes in value is not allowed // at position where '?' is not // present if (v1 != a1 && v1 != '?' ) return false ; if (v2 != a2 && v2 != '?' ) return false ; return true ; } // Function checks whether // the absolute difference // between hour and minute // value is within [L, R] bool inRange( int hh, int mm, int L, int R) { int a = abs (hh - mm); // Checks if the difference is outside // the give range if (a < L || a > R) return false ; return true ; } // Displays time in proper // 24-hour format void displayTime( int hh, int mm) { if (hh > 10) cout << hh << ":" ; else if (hh < 10) cout << "0" << hh << ":" ; if (mm > 10) cout << mm << endl; else if (mm < 10) cout << "0" << mm << endl; } // Function find the desired // value of time whose difference // lies in the range [L, R] void maximumTimeWithDifferenceInRange( string str, int L, int R) { int i, j; int h1, h2, m1, m2; // Decrease hour value from 23 to 0 for (i = 23; i >= 0; i--) { h1 = i % 10; h2 = i / 10; // Check if the hour value is valid // if not valid then no need to change // minute value, since time will still // remain in valid, to check hour value // flag is set to 1. if (!isValid(h1 + '0' , h2 + '0' , str, 1)) { continue ; } // Decrease minute value from 59 to 0 for (j = 59; j >= 0; j--) { m1 = j % 10; m2 = j / 10; // Check if the minute value is valid, // if not valid then skip the current // iteration, to check 'minute' value // flag is set to 0. if (!isValid(m1 + '0' , m2 + '0' , str, 0)) { continue ; } if (inRange(i, j, L, R)) { displayTime(i, j); return ; } } } if (inRange(i, j, L, R)) displayTime(i, j); else cout << "-1" << endl; } // Driver code int main() { // Input time string timeValue = "??:??" ; // Difference range int L = 20, R = 39; maximumTimeWithDifferenceInRange( timeValue, L, R); return 0; } |
Java
// Java program for // the above approach import java.util.*; class GFG { // Function checks whether // given time is correct static boolean isValid( char a1, char a2, String str, int flag) { char v1, v2; // To check minute value of time if (flag == 0 ) { v1 = str.charAt( 4 ); v2 = str.charAt( 3 ); } else { // To check hour value of time v1 = str.charAt( 1 ); v2 = str.charAt( 0 ); } // Changes in value is not allowed // at position where '?' is not // present if (v1 != a1 && v1 != '?' ) return false ; if (v2 != a2 && v2 != '?' ) return false ; return true ; } // Function checks whether // the absolute difference // between hour and minute // value is within [L, R] static boolean inRange( int hh, int mm, int L, int R) { int a = Math.abs(hh - mm); // Checks if the difference // is outside the give range if (a < L || a > R) return false ; return true ; } // Displays time in proper // 24-hour format static void displayTime( int hh, int mm) { if (hh > 10 ) System.out.print(hh + ":" ); else if (hh < 10 ) System.out.print( "0" + hh + ":" ); if (mm > 10 ) System.out.println(mm); else if (mm < 10 ) System.out.println( "0" + mm); } // Function find the desired // value of time whose difference // lies in the range [L, R] static void maximumTimeWithDifferenceInRange(String str, int L, int R) { int i = 0 , j = 0 ; int h1, h2, m1, m2; // Decrease hour value // from 23 to 0 for (i = 23 ; i >= 0 ; i--) { h1 = i % 10 ; h2 = i / 10 ; // Check if the hour value // is valid if not valid // then no need to change // minute value, since time // will still remain in valid, // to check hour value // flag is set to 1. if (!isValid(( char )h1, ( char )h2, str, 1 )) { continue ; } // Decrease minute value // from 59 to 0 for (j = 59 ; j >= 0 ; j--) { m1 = j % 10 ; m2 = j / 10 ; // Check if the minute value // is valid, if not valid // then skip the current // iteration, to check // 'minute' value // flag is set to 0. if (!isValid(( char )m1, ( char )m2, str, 0 )) { continue ; } if (inRange(i, j, L, R)) { displayTime(i, j); return ; } } } if (inRange(i, j, L, R)) displayTime(i, j); else System.out.println( "-1" ); } // Driver code public static void main(String[] args) { // Input time String timeValue = "??:??" ; // Difference range int L = 20 , R = 39 ; maximumTimeWithDifferenceInRange(timeValue, L, R); } } // This code is contributed by Chitranayal |
Python3
# Python3 program for the above approach # Function checks whether # given time is correct def isValid(a1, a2, strr, flag): v1, v2 = 0 , 0 # To check minute value of time if (flag = = 0 ): v1 = strr[ 4 ] v2 = strr[ 3 ] else : # To check hour value of time v1 = strr[ 1 ] v2 = strr[ 0 ] # Changes in value is not allowed # at position where '?' is not # present if (v1 ! = a1 and v1 ! = '?' ): return False if (v2 ! = a2 and v2 ! = '?' ): return False return True # Function checks whether # the absolute difference # between hour and minute # value is within [L, R] def inRange(hh, mm, L, R): a = abs (hh - mm) # Checks if the difference is # outside the give range if (a < L or a > R): return False return True # Displays time in proper # 24-hour format def displayTime(hh, mm): if (hh > 10 ): print (hh, end = ":" ) elif (hh < 10 ): print ( "0" , hh, end = ":" ) if (mm > 10 ): print (mm) elif (mm < 10 ): print ( "0" , mm) # Function find the desired # value of time whose difference # lies in the range [L, R] def maximumTimeWithDifferenceInRange(strr, L, R): i, j = 0 , 0 h1, h2, m1, m2 = 0 , 0 , 0 , 0 # Decrease hour value from 23 to 0 for i in range ( 23 , - 1 , - 1 ): h1 = i % 10 h2 = i / / 10 # Check if the hour value is valid # if not valid then no need to change # minute value, since time will still # remain in valid, to check hour value # flag is set to 1. if ( not isValid( chr (h1), chr (h2), strr, 1 )): continue # Decrease minute value from 59 to 0 for j in range ( 59 , - 1 , - 1 ): m1 = j % 10 m2 = j / / 10 # Check if the minute value is valid, # if not valid then skip the current # iteration, to check 'minute' value # flag is set to 0. if ( not isValid( chr (m1), chr (m2), strr, 0 )): continue if (inRange(i, j, L, R)): displayTime(i, j) return if (inRange(i, j, L, R)): displayTime(i, j) else : print ( - 1 ) # Driver code # Input time timeValue = "??:??" # Difference range L = 20 R = 39 maximumTimeWithDifferenceInRange(timeValue, L, R) # This code is contributed by mohit kumar 29 |
C#
// C# program for the above approach using System; class GFG{ // Function checks whether // given time is correct static bool isValid( char a1, char a2, string str, int flag) { char v1, v2; // To check minute value of time if (flag == 0) { v1 = str[4]; v2 = str[3]; } else { // To check hour value of time v1 = str[1]; v2 = str[0]; } // Changes in value is not allowed // at position where '?' is not // present if (v1 != a1 && v1 != '?' ) { return false ; } if (v2 != a2 && v2 != '?' ) { return false ; } return true ; } // Function checks whether // the absolute difference // between hour and minute // value is within [L, R] static bool inRange( int hh, int mm, int L, int R) { int a = Math.Abs(hh - mm); // Checks if the difference // is outside the give range if (a < L || a > R) { return false ; } return true ; } // Displays time in proper // 24-hour format static void displayTime( int hh, int mm) { if (hh > 10) { Console.Write(hh + ":" ); } else if (hh < 10) { Console.Write( "0" + hh + ":" ); } if (mm > 10) { Console.Write(mm); } else if (mm < 10) { Console.Write( "0" + mm); } } // Function find the desired // value of time whose difference // lies in the range [L, R] static void maximumTimeWithDifferenceInRange( string str, int L, int R) { int i = 0, j = 0; int h1, h2, m1, m2; // Decrease hour value // from 23 to 0 for (i = 23; i >= 0; i--) { h1 = i % 10; h2 = i / 10; // Check if the hour value // is valid if not valid // then no need to change // minute value, since time // will still remain in valid, // to check hour value // flag is set to 1. if (!isValid(( char )h1, ( char )h2, str, 1)) { continue ; } // Decrease minute value // from 59 to 0 for (j = 59; j >= 0; j--) { m1 = j % 10; m2 = j / 10; // Check if the minute value // is valid, if not valid // then skip the current // iteration, to check // 'minute' value // flag is set to 0. if (!isValid(( char )m1, ( char )m2, str, 0)) { continue ; } if (inRange(i, j, L, R)) { displayTime(i, j); return ; } } } if (inRange(i, j, L, R)) { displayTime(i, j); } else { Console.WriteLine( "-1" ); } } // Driver code static public void Main() { // Input time string timeValue = "??:??" ; // Difference range int L = 20, R = 39; maximumTimeWithDifferenceInRange(timeValue, L, R); } } // This code is contributed by avanitrachhadiya2155 |
Javascript
<script> // Javascript program for // the above approach // Function checks whether // given time is correct function isValid(a1, a2, str, flag) { let v1, v2; // To check minute value of time if (flag == 0) { v1 = str[4]; v2 = str[3]; } else { // To check hour value of time v1 = str[1]; v2 = str[0]; } // Changes in value is not allowed // at position where '?' is not // present if (v1 != a1 && v1 != '?' ) return false ; if (v2 != a2 && v2 != '?' ) return false ; return true ; } // Function checks whether // the absolute difference // between hour and minute // value is within [L, R] function inRange(hh,mm,L,R) { let a = Math.abs(hh - mm); // Checks if the difference // is outside the give range if (a < L || a > R) return false ; return true ; } // Displays time in proper // 24-hour format function displayTime(hh,mm) { if (hh > 10) document.write(hh + ":" ); else if (hh < 10) document.write( "0" + hh + ":" ); if (mm > 10) document.write(mm+ "<br>" ); else if (mm < 10) document.write( "0" + mm+ "<br>" ); } // Function find the desired // value of time whose difference // lies in the range [L, R] function maximumTimeWithDifferenceInRange(str,L,R) { let i = 0, j = 0; let h1, h2, m1, m2; // Decrease hour value // from 23 to 0 for (i = 23; i >= 0; i--) { h1 = i % 10; h2 = Math.floor(i / 10); // Check if the hour value // is valid if not valid // then no need to change // minute value, since time // will still remain in valid, // to check hour value // flag is set to 1. if (!isValid(String.fromCharCode(h1),String.fromCharCode(h2), str, 1)) { continue ; } // Decrease minute value // from 59 to 0 for (j = 59; j >= 0; j--) { m1 = j % 10; m2 = Math.floor(j / 10); // Check if the minute value // is valid, if not valid // then skip the current // iteration, to check // 'minute' value // flag is set to 0. if (!isValid(String.fromCharCode(m1), String.fromCharCode(m2), str, 0)) { continue ; } if (inRange(i, j, L, R)) { displayTime(i, j); return ; } } } if (inRange(i, j, L, R)) displayTime(i, j); else document.write( "-1<br>" ); } // Driver code // Input time let timeValue = "??:??" ; // Difference range let L = 20, R = 39; maximumTimeWithDifferenceInRange(timeValue, L, R); // This code is contributed by unknown2108 </script> |
23:59
Time Complexity: O(1)
Auxiliary Space: O(1)
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