Ephraimiyanda

Question

Approach

1. check if the length of the string is greater than 1. if its not we return false.
2. Make the string an array and loop through it.
3. Create a stack and store opening brackets inside.
4. Check if the next closing bracket is equal to the last element in stack. if it is i return true else i return false.

Complexity

• Time complexity: O(N)

Ephraimiyanda

Approach

I used the double multiplication (**) method for typescript

Complexity

• Time complexity:O(1)

• Space complexity:O(1)

Code

Ephraimiyanda

Question

Approach

To get the area between poles i created two pointers for the pair of poles that moves inward while calculating the areas between the poles also while limiting the height to be used for calculating the area to the shorter pole.

Complexity

• Time complexity: O(N)

• Space complexity: O(1)

Code

Ephraimiyanda

Question

Approach

I looped through the numbersand checked in pairs if they were duplicates.

Complexity

• Time complexity: O(N log N)

• Space complexity: O(N)

Code

Ephraimiyanda

Question

Approach

I looped through the string and sliced it by the length of the needle.if the word formed is the same as the needle then i return the index of the first letter of the word from the haystack.

Complexity

• Time complexity: O(M∗N)

• Space complexity: O(1)

Code

Ephraimiyanda

Approach

1.I first get the smallest word in the array of words. 2.A tracker to keep track of the length of the prefix is set. 3.A loop to go through the array and a loop to check whether each word starts with the prefix is created. if any of the words do not contain the prefix the prefix is reduced till all the words contaiin the longest prefix.

Complexity

• Time complexity: O(S)

• Space complexity: O(1)

Ephraimiyanda

Question

Approach

I turnedthe string into an array and filtered for words with characters greater than 0 to remove spaces and then retrieved the length of the last word.

Complexity

• Time complexity: O(N)

• Space complexity: O(N)

Code

Ephraimiyanda

Approach

i looped through thr array checking to see which number was the target and return the index and if the target did not exist among the numbers i return -1.

Complexity

• Time complexity: O(N) The time complexity for the question is supposed to be solved using O(log n). So this will need a revisit to be solved properly

• Space complexity: O(1)

Code

Ephraimiyanda

Approach

i loop through the numbers and assign a pointer for the next location of a val. if there is no val i overwrite the non val element and return the length on the non value elements.

Complexity

• Time complexity: O(N)

• Space complexity: O(1)

Code

Ephraimiyanda

Approach

setting the limit for the integer was a priority before turning the number into an array to be able to reverse it.Afterwards logic to check if the original integer was a negative or positive integer so that the reversed will be negative or positive.

Complexity

• Time complexity: O(N)

• Space complexity: O(N)

Code