summary
- Given an integer array, return true if there are three consecutive odd numbers in the array, otherwise return false.
approach
- The solution uses a counter to keep track of the number of consecutive odd numbers
- It iterates over the array, incrementing the counter whenever it encounters an odd number and resetting it whenever it encounters an even number
- If the counter reaches 3, it immediately returns true
- If it iterates over the entire array without finding three consecutive odd numbers, it returns false.
complexity
- O(n) where n is the number of elements in the array
explain
- The solution has a time complexity of O(n) because it makes a single pass over the array
- The space complexity is O(1) because it uses a constant amount of space to store the counter.
Solution Code:
class Solution {
public:
bool threeConsecutiveOdds(vector& arr) {
int cnt = 0;
for(int i = 0 ; i < arr.size(); i++){
if(arr[i] & 1){
cnt++;
} else {
cnt = 0;
}
if (cnt >= 3) return true;
}
return false;
}
};
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