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Add given n time durations
  • Last Updated : 07 Jun, 2019

Given n time durations in the form of MM : SS, where MM denotes minutes and SS denotes the seconds. The task is to add all the time duration and output answer in the form of HH : MM : SS.

Examples:

Input : n = 5
        duration1 = 0 : 45
        duration1 = 2 : 31
        duration1 = 3 : 11
        duration1 = 2 : 27
        duration1 = 1 : 28
Output : 0 : 10 : 22
Initially, sum = 0
On adding duration 1, sum = 0 hour 0 minutes 45 seconds.
On adding duration 2, sum = 0 hour 3 minutes 16 seconds.
On adding duration 3, sum = 0 hour 6 minutes 27 seconds.
On adding duration 4, sum = 0 hour 8 minutes 54 seconds.
On adding duration 5, sum = 0 hour 10 minutes 22 seconds

The idea is to convert all the given time duration to seconds. Once we know the duration in seconds, we can calculate the sum of durations in seconds.
In order to get the number of hours, we have to divide the total duration in seconds by 3600. The remainder is used to calculate the number of minutes and seconds. By dividing the remainder with 60 we get the number of minutes, and the remainder of that division is the number of seconds.

Below is the implementation of this approach:

C++




// CPP Program to find the sum of all duration 
// in the form of MM : SS.
#include <bits/stdc++.h>
using namespace std;
  
// Print sum of all duration.
void printSum(int m[], int s[], int n)
{
    int total = 0;
  
    // finding total seconds
    for (int i = 0; i < n; i++) {
        total += s[i];
        total += (m[i] * 60);
    }
  
    // print the hours.
    cout << total / 3600 << " : ";
    total %= 3600;
  
    // print the minutes.
    cout << total / 60 << ": ";
    total %= 60;
  
    // print the hours.
    cout << total << endl;
}
  
// Driven Program
int main()
{
    int m[] = { 0, 2, 3, 2, 1 };
    int s[] = { 45, 31, 11, 27, 28 };
    int n = sizeof(m)/sizeof(m[0]);
    printSum(m, s, n);
    return 0;
}


Java




// Java Program to find the
// sum of all duration in
// the form of MM : SS.
class GFG
{
      
    // Print sum of all duration.
    static void printSum(int m[], 
                    int s[], int n)
    {
        int total = 0;
      
        // finding total seconds
        for (int i = 0; i < n; i++)
        {
            total += s[i];
            total += (m[i] * 60);
        }
      
        // print the hours.
        System.out.print(total / 3600 + " : ");
        total %= 3600;
      
        // print the minutes.
        System.out.print(total / 60 +": ");
        total %= 60;
      
        // print the hours.
        System.out.println(total);
    
      
// Driver code 
public static void main (String[] args)
{
    int m[] = {0, 2, 3, 2, 1 };
    int s[] = { 45, 31, 11, 27, 28 };
    int n = m.length;
      
    printSum(m, s, n);
}
  
}
  
// This code is contributed by Anant Agarwal.


Python3




# Python3 code to find the sum of
# all duration in the form of MM : SS.
  
# Print sum of all duration.
def printSum (m, s, n ):
    total = 0
      
    # finding total seconds
    for i in range(n):
        total += s[i]
        total += (m[i] * 60)
      
    # print the hours.
    print(int(total / 3600) , end= " : ")
    total %= 3600
      
    # print the minutes.
    print(int(total / 60) , end=": ")
    total %= 60
      
    # print the hours.
    print(int(total))
  
# Driven Code
m = [ 0, 2, 3, 2, 1 ]
s = [ 45, 31, 11, 27, 28 ]
n = len(m)
printSum(m, s, n)
  
# This code is contributed by "Sharad_Bhardwaj".


C#




// C# Program to find the
// sum of all duration in
// the form of MM : SS.
using System;
  
class GFG
{
      
    // Print sum of all duration.
    static void printSum(int []m, 
                    int []s, int n)
    {
        int total = 0;
      
        // finding total seconds
        for (int i = 0; i < n; i++)
        {
            total += s[i];
            total += (m[i] * 60);
        }
      
        // print the hours.
        Console.Write(total / 3600 + " : ");
        total %= 3600;
      
        // print the minutes.
        Console.Write(total / 60 +": ");
        total %= 60;
      
        // print the hours.
        Console.Write(total);
    
      
    // Driver code 
    public static void Main ()
    {
        int []m = {0, 2, 3, 2, 1 };
        int []s = { 45, 31, 11, 27, 28 };
        int n = m.Length;
          
        printSum(m, s, n);
    }
}
  
// This code is contributed by vt_m.


PHP




<?php
// PHP Program to find the 
// sum of all duration 
// in the form of MM : SS.
  
// Print sum of all duration.
function printSum($m, $s, $n)
{
    $total = 0;
  
    // finding total seconds
    for ($i = 0; $i < $n; $i++) 
    {
        $total += $s[$i];
        $total += ($m[$i] * 60);
    }
  
    // print the hours.
    echo floor($total / 3600) , " : ";
    $total %= 3600;
  
    // print the minutes.
    echo floor($total / 60) ,": ";
    $total %= 60;
  
    // print the hours.
    echo floor($total) ;
      
}
  
    // Driver Code
    $m = array(0, 2, 3, 2, 1);
    $s= array(45, 31, 11, 27, 28);
    $n = count($m);
    printSum($m, $s, $n);
  
// This code is contributed by anuj_67.
?>



Output:

0 : 10: 22



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