In this approach, I traverse the grid to count the occurrences of each number using a dictionary. Then, I iterate through the expected range [1, n^2] to identify the repeated and missing numbers.
Ifihanagbara Olusheye Ifihan
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I observe that the pattern of growth follows a square expansion. At each step, new layers of cells are added around the previous configuration.
The formula for the total number of colored cells after n minutes is derived as: Total colored cells = n^2 + (n-1)^2
I used a greedy approach by repeatedly checking the remainder when dividing n by 3. If at any step, the remainder is 2, it means that n cannot be represented as a sum of distinct powers of three. Otherwise, I keep dividing n by 3 until it becomes zero.
class Solution:
def checkPowersOfThree(self, n: int) -> bool:I iterate through nums and categorize elements into three lists: less_than, equal_to, and greater_than. This shows that all elements less than pivot appear first, followed by elements equal to pivot, then elements greater than pivot.
The relative order of elements in each category is preserved as required.
For my approach, I use a dictionary to store the sum of values for each unique id. I iterate through both nums1 and nums2, adding their values to the dictionary. After processing both arrays, I convert the dictionary into a sorted list of lists.
class Solution:In my approach, I first iterate through the array and check if two consecutive elements are equal. If they are, I double the value of the first element and set the second to zero. After applying all operations sequentially, I filter out non-zero elements and store them in a new list.
Finally, I append the necessary number of zeros to maintain the original array size.
My approach was using a hash map index_map to map each value in the array. Then I initialized a 2D dynamic programming table (dp) where dp[j][i] stores the length of the longest Fibonacci-like subsequence ending with elements at indices j and i. All values were initially set to 2, representing the minimum length of a valid Fibonacci sequence.
During iteration over all pairs of indices (i, j) with i > j, I calculated the potential previous element as arr[i] - arr[j]. If this element existed in the array with an index less than j, it indicated a valid extension of the Fibonacci sequence. I updated dp[j][i] accordingly and tracked the maximum length found using max_length.
Finally, I returned max_length if it was at least 3, indicating a valid Fibonacci-like subsequence.
I first initialized variables to track the maximum and minimum sums as well as the current sums while iterating through the array. At each step, I updated the current maximum and minimum sums by either extending the existing subarray or starting fresh with the current element. At the same time, I maintained the global maximum and minimum sums observed during the traversal.
Finally, I returned the greater value between the maximum sum and the absolute value of the minimum sum.
I started by using two counters: even_count and odd_count. First, I set even_count to 1 to account for an empty prefix sum, and odd_count to 0. As I iterated through the array, I accumulated the prefix sum and checked whether it was odd or even. When the prefix sum was even, I added the odd_count to the result since combining an even prefix sum with a prior odd prefix would result in an odd subarray sum. Conversely, if the prefix sum was odd, I added the even_count to the result, on the fact that an odd prefix sum paired with a prior even prefix sum also yields an odd subarray sum.
To handle large outputs, I kept the result modulo (10^9 + 7) throughout the process.