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author | Jack Sangdahl <[email protected]> | 2025-02-24 20:14:35 -0700 |
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committer | Jack Sangdahl <[email protected]> | 2025-02-24 20:14:35 -0700 |
commit | 658556f3d4eaa1cdefd7009e7b6ccb3cabcd31ac (patch) | |
tree | 789c0f7874ff9d86e0f11e16daa5e15f0ed73e44 | |
parent | f86f165d87dacea43fb73a7f4106b3e3dbc3d83e (diff) | |
download | leetcode-658556f3d4eaa1cdefd7009e7b6ccb3cabcd31ac.tar.gz leetcode-658556f3d4eaa1cdefd7009e7b6ccb3cabcd31ac.zip |
1524. number of subarrays with odd sum
-rw-r--r-- | number_subarr_odd_sum.c | 24 | ||||
-rw-r--r-- | number_subarr_odd_sum.py | 28 |
2 files changed, 52 insertions, 0 deletions
diff --git a/number_subarr_odd_sum.c b/number_subarr_odd_sum.c new file mode 100644 index 0000000..bcc1569 --- /dev/null +++ b/number_subarr_odd_sum.c @@ -0,0 +1,24 @@ +// 1524. Number of Sub-arrays With Odd Sum +#include "leetcode.h" + +/* + * given an array of integers 'arr', return the number of subarrays with an odd + * sum. since the answer can be very large, return it modulo 10e9+7. + */ + +int numOfSubarrays(int *arr, int arrSize) { + int curr = 0, ans = 0, cnt[] = {1, 0}, mod = 1e9 + 7; + for (int i = 0; i < arrSize; i++) { + curr ^= arr[i] & 1; + ans = (ans + cnt[1 - curr]) % mod; + cnt[curr]++; + } + return ans; +} + +int main() { + int a1[] = {1, 3, 5}, a2[] = {2, 4, 6}, a3[] = {1, 2, 3, 4, 5, 6, 7}; + printf("%d\n", numOfSubarrays(a1, ARRAY_SIZE(a1))); // expect: 4 + printf("%d\n", numOfSubarrays(a2, ARRAY_SIZE(a2))); // expect: 0 + printf("%d\n", numOfSubarrays(a3, ARRAY_SIZE(a3))); // expect: 16 +} diff --git a/number_subarr_odd_sum.py b/number_subarr_odd_sum.py new file mode 100644 index 0000000..cf00796 --- /dev/null +++ b/number_subarr_odd_sum.py @@ -0,0 +1,28 @@ +# 1524. Number of Sub-arrays With Odd Sum + +""" +given an array of integers 'arr', return the number of subarrays with an odd +sum. since the answer can be very large, return it modulo 10e9+7. +""" + + +class Solution(object): + def numOfSubarrays(self, arr): + """ + :type arr: List[int] + :rtype: int + """ + cnt = [1, 0] + curr, ans = 0, 0 + for i in arr: + curr ^= i & 1 + ans += cnt[1 - curr] + cnt[curr] += 1 + return ans % (10**9 + 7) + + +if __name__ == "__main__": + obj = Solution() + print(obj.numOfSubarrays(arr=[1, 3, 5])) + print(obj.numOfSubarrays(arr=[2, 4, 6])) + print(obj.numOfSubarrays(arr=[1, 2, 3, 4, 5, 6, 7])) |