Pairs of Songs With Total Durations Divisible by 60 - LeetCode
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
- code TLE O(n^2)
class Solution:
def numPairsDivisibleBy60(self, time: List[int]) -> int:
ct = 0
n = len(time)
for i in range(n-1):
for j in range(i+1, n):
if (time[i] + time[j]) % 60 == 0: ct += 1
return ct
- code
class Solution:
def numPairsDivisibleBy60(self, time: List[int]) -> int:
ct = 0
needs = defaultdict(int) # time needed, count of need
for t in time:
t %= 60
if t == 0:
ct += needs[60]
elif t in needs:
ct += needs[t]
needs[60 - t] += 1
return ct
- code
class Solution {
public int numPairsDivisibleBy60(int[] time) {
int[] need = new int[60];
int count = 0;
for(int t: time){
int rem = t % 60;
count += need[rem];
if (rem == 0) need[0]++;
else need[60 - rem]++;
}
return count;
}
}