This solution has passed all HackerRank test cases, so I presume it's correct. Sample test cases: [3 3 9 9 5] mod 7 -> 6 [1 5 9] mod 5 -> 4 [1 2 3] mod 2 -> 1 [2 2 2 3] mod 7 -> 6
- [3 3 9 9 5] mod 7 → 6
- [1 5 9] mod 5 → 4
- [1 2 3] mod 2 → 1
- [2 2 2 3] mod 7 → 6
This solution has passed all HackerRank test cases, so I presume it's correct. Sample test cases: [3 3 9 9 5] mod 7 -> 6 [1 5 9] mod 5 -> 4 [1 2 3] mod 2 -> 1 [2 2 2 3] mod 7 -> 6
This solution has passed all HackerRank test cases, so I presume it's correct. Sample test cases:
- [3 3 9 9 5] mod 7 → 6
- [1 5 9] mod 5 → 4
- [1 2 3] mod 2 → 1
- [2 2 2 3] mod 7 → 6
This solution has passed all HackerRank test cases, so I presume it's correct. Sample test cases: [3 3 9 9 5] mod 7 -> 6 [1 5 9] mod 5 -> 4 [1 2 3] mod 2 -> 1 [2 2 2 3] mod 7 -> 6
This solution has passed all HackerRank test cases, so I presume it's correct. Sample test cases: [3 3 9 9 5] mod 7 -> 6 [1 5 9] mod 5 -> 4 [1 2 3] mod 2 -> 1 [2 2 2 3] mod 7 -> 6
A. The solution comes from a subarray that starts from the beginning;beginning, as in arr = [2, 2, 2, 3], M = 7 case;
B. Several subarrays that start from the beginning can have the same modulo, and there is no need to study all of them, only the shortest and the longest.
A. The solution comes from a subarray that starts from the beginning;
B. Several subarrays that start from the beginning can have the same modulo, and there is no need to study all of them, only the shortest and the longest.
A. The solution comes from a subarray that starts from the beginning, as in arr = [2, 2, 2, 3], M = 7 case;
B. Several subarrays that start from the beginning can have the same modulo, and there is no need to study all of them, only the shortest and the longest.