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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[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
Companies:
Goldman Sachs
Related Topics:
Array
Solution 1.
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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; } } ############ class Solution { public int numPairsDivisibleBy60(int[] time) { int[] cnt = new int[501]; int ans = 0; for (int t : time) { int s = 60; for (int i = 0; i < 17; ++i) { if (s - t >= 0 && s - t < cnt.length) { ans += cnt[s - t]; } s += 60; } cnt[t]++; } return ans; } }
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// 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[60] = {}, ans = 0; for (int n : A) { n %= 60; ans += cnt[(60 - n) % 60]; cnt[n]++; } return ans; } };
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class Solution: def numPairsDivisibleBy60(self, time: List[int]) -> int: cnt = defaultdict(int) ans = 0 for t in time: s = 60 for _ in range(17): ans += cnt[s - t] s += 60 cnt[t] += 1 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
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func numPairsDivisibleBy60(time []int) (ans int) { cnt := [501]int{} for _, t := range time { s := 60 for i := 0; i < 17; i++ { if s-t >= 0 && s-t < 501 { ans += cnt[s-t] } s += 60 } cnt[t]++ } return }
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function numPairsDivisibleBy60(time: number[]): number { const cnt: number[] = new Array(60).fill(0); for (const t of time) { ++cnt[t % 60]; } let ans = 0; for (let x = 1; x < 30; ++x) { ans += cnt[x] * cnt[60 - x]; } ans += (cnt[0] * (cnt[0] - 1)) / 2; ans += (cnt[30] * (cnt[30] - 1)) / 2; return ans; }