Problems
November 7, 2022
Maximum 69 number
We will iterate over all the digits of the number and check if the digit is 6. If it is, we will replace it with 9 and return the number.
Time complexity: O(n)
Space complexity:...
November 7, 2022
Wiggle sort II
We will sort the array and then split it into two halves. We will iterate over the first half and the second half and insert the elements into the result array. Time complexity:...
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Number of subsequences that satisfy the given sum condition
We will sort the array and use two pointers to find the number of subsequences that satisfy the given sum condition.
Time complexity: O(nlogn)
Space complexity: O(1)
November 7, 2022
Array with elements not equal to average of neighbors
We will sort the array. Then take 2 elements in pair at a time, swap their positions and then return it after the loop is finished.
Time complexity: O(nlogn)
Space complexity:...
November 6, 2022
Sort integers by the number of 1 bits
We will iterate over all the numbers in of the array and count the 1 bits and store it in a map. Then we will sort the map by the value and return the keys.
Time complexity: O(nlogn)...
November 6, 2022
Design an ordered stream
We will use a list to store the values and a pointer to keep track of the next index to be filled. We will return the values from the pointer to the index of the current value. Time complexity:...
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Check completeness of a binary tree
We will use BFS for level order traversal. We will add children to the BFS queue, if it's a complete tree, we should not have any node once we encounter a null node. Time complexity:...
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Jewels and stones
We will count all the distinct jewels and store it in a set. Then we will iterate over all the stones and check if the stone is a jewel. If it is, we will increment the count. We can also count...
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Orderly queue
For k>1 (we just check for k=2), any two adjacent characters can be swapped: abXYZ -> abXYZ -> aXYZb -> XYZba -> YZbaX -> ZbaXY -> baXYZ. If ab are not in the beginning of the string, i.e., DEabF, we...
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Minimum number of vertices to reach all nodes
We can just return all nodes with on incoming edge. Necesssary condition for this to work is to have all nodes with no in-degree must in the final result, because they can not be reached from. All...
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