In a list of songs, the i-th song has a duration of time[i] seconds.
Return the number of pairs of songs for which their total duration in seconds is divisible by 60. Formally, we want the number of indices i, j such that i < j with (time[i] + time[j]) % 60 == 0.
Example 1:
Input: [30,20,150,100,40]
Output: 3
Explanation: Three pairs have a total duration divisible by 60:
(time[0] = 30, time[2] = 150): total duration 180
(time[1] = 20, time[3] = 100): total duration 120
(time[1] = 20, time[4] = 40): total duration 60
Example 2:
Input: [60,60,60]
Output: 3
Explanation: All three pairs have a total duration of 120, which is divisible by 60.
Note:
1 <= time.length <= 60000
1 <= time[i] <= 500
Solution:
using System;
using System.Collections.Generic;
using System.Text;
namespace LeetCode.AskGif.Easy.Array
{
public class NumPairsDivisibleBy60Soln
{
public int NumPairsDivisibleBy60(int[] time)
{
var map = new Dictionary<int, int>();
int count = 0;
int val = 0;
int diff = 0;
for (int i = 0; i < time.Length; i++)
{
val = time[i] % 60;
diff = (60 - val) % 60;
if (map.ContainsKey(diff))
{
count += map[diff];
}
if (map.ContainsKey(val))
{
map[val]++;
}
else
{
map.Add(val, 1);
}
}
return count;
}
}
}
Time Complexity: O(n)
Space Complexity: O(n)
Unit Tests:
using LeetCode.AskGif.Easy.Array;
using Microsoft.VisualStudio.TestTools.UnitTesting;
using System;
using System.Collections.Generic;
using System.Text;
namespace CodingUnitTest.Easy.Array
{
[TestClass]
public class NumPairsDivisibleBy60SolnTests
{
[TestMethod]
public void NumPairsDivisibleBy60Soln_First()
{
var arr = new int[] { 30, 20, 150, 100, 40 };
var expected = 3;
var res = new NumPairsDivisibleBy60Soln().NumPairsDivisibleBy60(arr);
Assert.AreEqual(expected, res);
}
[TestMethod]
public void NumPairsDivisibleBy60Soln_Second()
{
var arr = new int[] { 60, 60, 60 };
var expected = 3;
var res = new NumPairsDivisibleBy60Soln().NumPairsDivisibleBy60(arr);
Assert.AreEqual(expected, res);
}
}
}



