1010. Pairs of Songs With Total Durations Divisible by 60
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
# @lc code=start
using LeetCode
function num_pairs_divisible_by60(time::Vector{Int})
res = 0
dct = fill(0, 60)
for t in time
(t % 60 == 0) ? (dct[60] += 1) : (dct[t % 60] += 1)
end
for i in 1:29
res += dct[i] * dct[60 - i]
end
res += sum(1:(dct[30] - 1)) + sum(1:(dct[60] - 1))
return res
end
# @lc code=end
num_pairs_divisible_by60 (generic function with 1 method)
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