Given an array arr[][2] consisting of N pair of strings representing the starting and ending time (in 12 hours format) and a string P which represents the time of the meeting, the task is to find the count of intervals that contains the time P.
Examples:
Input: P = “12:01:AM”, arr[][2] = {{“12:00:AM”, “11:55:PM”}, {“12:01:AM”, “11:50:AM”}, {“12:30:AM”, “12:00:PM”}, {“11:57:AM”, “11:59:PM”}}
Output: 2
Explanation: The time P lies in the first and second time intervals.Input: P = “12:01:AM”, arr[][2] = {{“09:57:AM”, “12:00:PM”} }
Output: 0
Explanation: No interval contains the time P.
Approach: The idea is to first convert all the times from 12-hour format into 24-hour format and then compare the range with Time P. Follow the steps below to solve the problem:
- First, convert all the times of 12 hours format to 24 hours format and then store the integer value of 24 -hours time format.
- Initialize a variable, say ans, to store the count of intervals in which P lies.
- Traverse the array arr[] and increment the count of ans by 1 if (P ? L && P ? R) or (P ? R && P ? L).
- Finally, after completing the above steps, print ans.
Below is the implementation of the above approach:
// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to convert a time in 24
// hour format to an equivalent integer
int convert(string str)
{
// Removes ":" at 3rd position
str.replace(2, 1, "");
// Calculate hours
int h1 = (int)str[1] - '0';
int h2 = (int)str[0] - '0';
int hh = (h2 * 10 + h1 % 10);
// Stores the time in 24 hours format
int time = 0;
// If time is in "AM"
if (str[5] == 'A') {
// If hh is equal to 12
if (hh == 12)
time += stoi(str.substr(2, 2));
else {
time += stoi(str.substr(0, 2));
}
}
// If time is in "PM"
else {
// If hh is equal to 12
if (hh == 12) {
time += stoi(str.substr(0, 4));
}
else {
time += stoi(str.substr(0, 4));
time += 1200;
}
}
// Return time
return time;
}
// Function to count number
// of intervals in which p lies
int countOverlap(string arr[][2],
int n, string p)
{
// Stores the count
int ans = 0;
// Stores the integer value of
// 24 hours time format of P
int M = convert(p);
// Traverse the array
for (int i = 0; i < n; i++) {
// Stores the integer value of
// 24 hours time format of arr[i][0]
int L = convert(arr[i][0]);
// Stores the integer value of
// 24 hours time format of arr[i][1]
int R = convert(arr[i][1]);
// If M lies within the [L, R]
if ((L <= M && M <= R)
|| (M >= R && M <= L))
// Increment ans by 1
ans++;
}
// Return ans
return ans;
}
// Driver Code
int main()
{
string arr[][2] = { { "12:00:AM", "11:55:PM" },
{ "12:01:AM", "11:50:AM" },
{ "12:30:AM", "12:00:PM" },
{ "11:57:AM", "11:59:PM" } };
string P = "12:01:PM";
int N = sizeof(arr) / sizeof(arr[0]);
cout << countOverlap(arr, N, P) << endl;
}
// Java implementation of the above approach
import java.io.*;
class GFG
{
// Function to convert a time in 24
// hour format to an equivalent integer
static int convert(String str)
{
// Removes ":" at 3rd position
str = str.substring(0, 2) + str.substring(3);
// Calculate hours
int h1 = (int)str.charAt(1) - '0';
int h2 = (int)str.charAt(0) - '0';
int hh = (h2 * 10 + h1 % 10);
// Stores the time in 24 hours format
int time = 0;
// If time is in "AM"
if (str.charAt(5) == 'A') {
// If hh is equal to 12
if (hh == 12)
time += Integer.parseInt(
str.substring(2, 4));
else {
time += Integer.parseInt(
str.substring(0, 2));
}
}
// If time is in "PM"
else {
// If hh is equal to 12
if (hh == 12) {
time += Integer.parseInt(
str.substring(0, 4));
}
else {
time += Integer.parseInt(
str.substring(0, 4));
time += 1200;
}
}
// Return time
return time;
}
// Function to count number
// of intervals in which p lies
static int countOverlap(String arr[][], int n, String p)
{
// Stores the count
int ans = 0;
// Stores the integer value of
// 24 hours time format of P
int M = convert(p);
// Traverse the array
for (int i = 0; i < n; i++)
{
// Stores the integer value of
// 24 hours time format of arr[i][0]
int L = convert(arr[i][0]);
// Stores the integer value of
// 24 hours time format of arr[i][1]
int R = convert(arr[i][1]);
// If M lies within the [L, R]
if ((L <= M && M <= R) || (M >= R && M <= L))
// Increment ans by 1
ans++;
}
// Return ans
return ans;
}
// Driver Code
public static void main(String[] args)
{
String[][] arr
= new String[][] { { "12:00:AM", "11:55:PM" },
{ "12:01:AM", "11:50:AM" },
{ "12:30:AM", "12:00:PM" },
{ "11:57:AM", "11:59:PM" } };
String P = "12:01:PM";
int N = arr.length;
System.out.println(countOverlap(arr, N, P));
}
}
// This code is contributed by Dharanendra L V
# Python3 program to implement the approach
# Function to convert a time in 24
# hour format to an equivalent integer
def convert(str):
# Removes ":" at 3rd position
str = str.replace(":", "", 1)
# Calculate hours
hh = int(str[0:2])
# Stores the time in 24 hours format
time = 0
# If time is in "AM"
if str[-1] == "A":
# If hh is equal to 12
if hh == 12:
time += int(str[2:4])
else:
time += int(str[0:2])
# If time is in "PM"
else:
# If hh is equal to 12
if hh == 12:
time += int(str[0:4])
else:
time += int(str[0:4])
time += 1200
# Return time
return time
# Function to count number
# of intervals in which p lies
def countOverlap(arr, n, p):
# Stores the count
ans = 0
# Stores the integer value of
# 24 hours time format of P
M = convert(p)
# Traverse the array
for i in range(n):
# Stores the integer value of
# 24 hours time format of arr[i][0]
L = convert(arr[i][0])
# Stores the integer value of
# 24 hours time format of arr[i][1]
R = convert(arr[i][1])
# If M lies within the [L, R]
if L <= M <= R :
# Increment ans by 1
ans += 1
# Return ans
return ans
# Driver Code
arr = [["12:00:AM", "11:55:PM"],
["12:01:AM", "11:50:AM"],
["12:30:AM", "12:00:PM"],
["11:57:AM", "11:59:PM"]]
P = "12:01:PM"
N = len(arr)
print(countOverlap(arr, N, P))
// C# implementation of the above approach
using System;
class GFG{
// Function to convert a time in 24
// hour format to an equivalent integer
static int convert(String str)
{
// Removes ":" at 3rd position
str = str.Substring(0, 2) + str.Substring(3);
// Calculate hours
int h1 = (int)str[1] - '0';
int h2 = (int)str[0] - '0';
int hh = (h2 * 10 + h1 % 10);
// Stores the time in 24 hours format
int time = 0;
// If time is in "AM"
if (str[5] == 'A')
{
// If hh is equal to 12
if (hh == 12)
time += Int32.Parse(
str.Substring(2, 2));
else
{
time += Int32.Parse(
str.Substring(0, 2));
}
}
// If time is in "PM"
else
{
// If hh is equal to 12
if (hh == 12)
{
time += Int32.Parse(
str.Substring(0, 4));
}
else
{
time += Int32.Parse(
str.Substring(0, 4));
time += 1200;
}
}
// Return time
return time;
}
// Function to count number
// of intervals in which p lies
static int countOverlap(String [,]arr, int n,
String p)
{
// Stores the count
int ans = 0;
// Stores the integer value of
// 24 hours time format of P
int M = convert(p);
// Traverse the array
for(int i = 0; i < n; i++)
{
// Stores the integer value of
// 24 hours time format of arr[i,0]
int L = convert(arr[i,0]);
// Stores the integer value of
// 24 hours time format of arr[i,1]
int R = convert(arr[i,1]);
// If M lies within the [L, R]
if ((L <= M && M <= R) ||
(M >= R && M <= L))
// Increment ans by 1
ans++;
}
// Return ans
return ans;
}
// Driver Code
public static void Main(String[] args)
{
String[,] arr = new String[,]{
{ "12:00:AM", "11:55:PM" },
{ "12:01:AM", "11:50:AM" },
{ "12:30:AM", "12:00:PM" },
{ "11:57:AM", "11:59:PM" } };
String P = "12:01:PM";
int N = arr.GetLength(0);
Console.WriteLine(countOverlap(arr, N, P));
}
}
// This code is contributed by 29AjayKumar
// JS program to implement the approach
function convert(str) {
// Removes ":" at 3rd position
str = str.replace(":", "", 1);
// Calculate hours
const hh = parseInt(str.substring(0, 2));
// Stores the time in 24 hours format
let time = 0;
// If time is in "AM"
if (str.charAt(str.length - 1) === "A") {
// If hh is equal to 12
if (hh === 12) {
time += parseInt(str.substring(2, 4));
} else {
time += parseInt(str.substring(0, 2));
}
} else {
// If hh is equal to 12
if (hh === 12) {
time += parseInt(str.substring(0, 4));
} else {
time += parseInt(str.substring(0, 4));
time += 1200;
}
}
// Return time
return time;
}
function countOverlap(arr, n, p) {
// Stores the count
let ans = 0;
// Stores the integer value of
// 24 hours time format of P
const M = convert(p);
// Traverse the array
for (let i = 0; i < n; i++) {
// Stores the integer value of
// 24 hours time format of arr[i][0]
const L = convert(arr[i][0]);
// Stores the integer value of
// 24 hours time format of arr[i][1]
const R = convert(arr[i][1]);
// If M lies within the [L, R]
if (L <= M && M <= R) {
// Increment ans by 1
ans++;
}
}
// Return ans
return ans;
}
const arr = [ ["12:00:AM", "11:55:PM"],
["12:01:AM", "11:50:AM"],
["12:30:AM", "12:00:PM"],
["11:57:AM", "11:59:PM"],
];
const P = "12:01:PM";
const N = arr.length;
console.log(countOverlap(arr, N, P));
// This code is contributed by phasing17
Output:
2
Time Complexity: O(N)
Auxiliary Space: O(1)