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.



  1. Since, maximum time value is desired, so we decrease ‘hour’ value from 23 to 0 and similarly ‘minute’ value from 59 to 0.
  2. After decreasing ‘hour’ and ‘minute’ value, we need to check if the ‘hour’ and ‘minute’ value is valid or not.
  3. Change in value of ‘hour’ and ‘minute’ is allowed only at those position which have ‘?’. Changes at other postion will make time value invalid. To ensure this we will call function isValid().
  4. Keep on decreasing ‘hour’ and ‘minute’ value untill a valid time is found whose difference lies in the range [ L, R ].
  5. If no valid time is found, then print “-1”.

Below is the implementation of the above approach:

C++

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// 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;
}

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Python3

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# 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

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Output:

23:59

Time Complexity: O(1)
Auxiliary Space: O(1)

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Improved By : mohit kumar 29, nidhi_biet

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