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PairOfSongsWithGivenTotalDuration.java
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PairOfSongsWithGivenTotalDuration.java
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package Leetcode;
import java.util.HashMap;
import java.util.Map;
/**
* @author kalpak
*
* You are given a list of songs where the ith 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: time = [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: time = [60,60,60]
* Output: 3
* Explanation: All three pairs have a total duration of 120, which is divisible by 60.
*
*
* Constraints:
*
* 1 <= time.length <= 6 * 104
* 1 <= time[i] <= 500
*
*/
public class PairOfSongsWithGivenTotalDuration {
public static int numPairsDivisibleBy60(int[] time) {
Map<Integer, Integer> map = new HashMap<>();
int result = 0;
for(int i : time) {
int otherTime = (60 - i % 60) % 60; // calculate the other time for the current time t.
if(map.containsKey(otherTime)) // if such a pair exists, add it to the result
result += map.get(otherTime);
map.put(i % 60, map.getOrDefault(i%60, 0) + 1); // add the current time to the map
}
return result;
}
public static void main(String[] args) {
int[] time = new int[]{30,20,150,100,40};
System.out.println(numPairsDivisibleBy60(time));
}
}