Formatted question description: https://leetcode.ca/all/1010.html

# 1010. Pairs of Songs With Total Durations Divisible by 60 (Medium)

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 = 30, time = 150): total duration 180
(time = 20, time = 100): total duration 120
(time = 20, time = 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

Companies:
Goldman Sachs

Related Topics:
Array

## Solution 1.

// OJ: https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int numPairsDivisibleBy60(vector<int>& A) {
int cnt = {}, ans = 0;
for (int n : A) {
n %= 60;
ans += cnt[(60 - n) % 60];
cnt[n]++;
}
return ans;
}
};


Java

• class Solution {
public int numPairsDivisibleBy60(int[] time) {
int length = time.length;
int[] remainder = new int[length];
for (int i = 0; i < length; i++)
remainder[i] = time[i] % 60;
Arrays.sort(remainder);
int beginIndex = 0;
while (beginIndex < length && remainder[beginIndex] == 0)
beginIndex++;
int pairs = beginIndex * (beginIndex - 1) / 2;
int low = beginIndex, high = length - 1;
while (low < high) {
int sum = remainder[low] + remainder[high];
if (sum < 60)
low++;
else if (sum > 60)
high--;
else {
if (remainder[low] == 30) {
int count = high - low + 1;
pairs += count * (count - 1) / 2;
break;
} else {
int lowEnd = low, highEnd = high;
int remainderLow = remainder[low], remainderHigh = remainder[high];
while (lowEnd < high && remainder[lowEnd] == remainderLow)
lowEnd++;
while (highEnd > low && remainder[highEnd] == remainderHigh)
highEnd--;
pairs += (lowEnd - low) * (high - highEnd);
low = lowEnd;
high = highEnd;
}
}
}
return pairs;
}
}

• // OJ: https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int numPairsDivisibleBy60(vector<int>& A) {
int cnt = {}, ans = 0;
for (int n : A) {
n %= 60;
ans += cnt[(60 - n) % 60];
cnt[n]++;
}
return ans;
}
};

• # 1010. Pairs of Songs With Total Durations Divisible by 60
# https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/

class Solution:
def numPairsDivisibleBy60(self, time: List[int]) -> int:
s = res = 0
mp = Counter()

for x in time:
target = (60 - x) % 60

res += mp[target]

mp[x % 60] += 1

return res