LeetCode 2501: Longest Square Streak in an Array
LeetCode 2501 Solution Explanation
Explanation:
To solve this problem, we can iterate through each element in the array and consider it as the starting point of a potential square streak. We then try to build the square streak by looking for the square of the current element in the remaining elements. If we find the square, we continue this process until we can no longer find the next square in the sequence.
Algorithmic Idea:
- Initialize a variable
longestStreak
to keep track of the maximum square streak length found so far. - Iterate through each element in the array.
- For each element
num[i]
, try to build a square streak starting fromnum[i]
. - Check if
num[i] * num[i]
exists innums
, if it does, increment the streak length by 1 and update the current element to the square found. - Continue this process until we can no longer find the next square in the sequence.
- Update
longestStreak
if the current streak is longer.
Time Complexity:
The time complexity of this approach is O(n^2) where n is the length of the input array nums
.
Space Complexity:
The space complexity of this approach is O(1) as we are using only a constant amount of extra space.
LeetCode 2501 Solutions in Java, C++, Python
class Solution {
public int longestSquareStreak(int[] nums) {
int longestStreak = -1;
for (int i = 0; i < nums.length; i++) {
int curr = nums[i];
int streak = 1;
while (true) {
int next = curr * curr;
boolean found = false;
for (int j = i + 1; j < nums.length; j++) {
if (nums[j] == next) {
streak++;
curr = nums[j];
found = true;
break;
}
}
if (!found) break;
}
longestStreak = Math.max(longestStreak, streak);
}
return longestStreak >= 2 ? longestStreak : -1;
}
}
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