"Use a representative of each, e.g. sort the string and add it to the value of a hashmap> where we put all the words that belong to the same anagram together."
Gaston B. - "Use a representative of each, e.g. sort the string and add it to the value of a hashmap> where we put all the words that belong to the same anagram together."See full answer
"Initialize left pointer: Set a left pointer left to 0.
Iterate through the array: Iterate through the array from left to right.
If the current element is not 0, swap it with the element at the left pointer and increment left.
Time complexity: O(n). The loop iterates through the entire array once, making it linear time.
Space complexity: O(1). The algorithm operates in-place, modifying the input array directly without using additional data structures.
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Avon T. - "Initialize left pointer: Set a left pointer left to 0.
Iterate through the array: Iterate through the array from left to right.
If the current element is not 0, swap it with the element at the left pointer and increment left.
Time complexity: O(n). The loop iterates through the entire array once, making it linear time.
Space complexity: O(1). The algorithm operates in-place, modifying the input array directly without using additional data structures.
"See full answer
"a. Sort the array elements.
b. take two pointers at index 0 and index Len-1;
c. if the sum at the two pointers is target; break and return the pair
d. if the sum is smaller, then move left pointer by 1
e. else move right pointer by 1;
run the logic till the target is met or right pointer crosses the left pointer."
Komal S. - "a. Sort the array elements.
b. take two pointers at index 0 and index Len-1;
c. if the sum at the two pointers is target; break and return the pair
d. if the sum is smaller, then move left pointer by 1
e. else move right pointer by 1;
run the logic till the target is met or right pointer crosses the left pointer."See full answer