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For the input string 'abcab', the first non-repeating character is ‘c’. As depicted the character ‘a’ repeats at index 3 and character ‘b’ repeats at index 4. Hence we return the character ‘c’ present at index 2.



You can only stack a box on top of another box if the dimensions of the 2-D base of the lower box ( both length and width ) are strictly larger than those of the 2-D base of the higher box.
You can rotate a box so that any side functions as its base. It is also allowed to use multiple instances of the same type of box. This means, a single type of box when rotated, will generate multiple boxes with different dimensions, which may also be included in stack building.

The height, Width, Length of the type of box will interchange after rotation.
No two boxes will have all three dimensions the same.
Don’t print anything, just return the height of the highest possible stack that can be formed.
Technical Round



I used dynamic programming to solve this length of palindromic subsequence in string



Consider below matrix of characters,
[ 'D', 'E', 'X', 'X', 'X' ]
[ 'X', 'O', 'E', 'X', 'E' ]
[ 'D', 'D', 'C', 'O', 'D' ]
[ 'E', 'X', 'E', 'D', 'X' ]
[ 'C', 'X', 'X', 'E', 'X' ]
If the given string is "CODE", below are all its occurrences in the matrix:
'C'(2, 2) 'O'(1, 1) 'D'(0, 0) 'E'(0, 1)
'C'(2, 2) 'O'(1, 1) 'D'(2, 0) 'E'(3, 0)
'C'(2, 2) 'O'(1, 1) 'D'(2, 1) 'E'(1, 2)
'C'(2, 2) 'O'(1, 1) 'D'(2, 1) 'E'(3, 0)
'C'(2, 2) 'O'(1, 1) 'D'(2, 1) 'E'(3, 2)
'C'(2, 2) 'O'(2, 3) 'D'(2, 4) 'E'(1, 4)
'C'(2, 2) 'O'(2, 3) 'D'(3, 3) 'E'(3, 2)
'C'(2, 2) 'O'(2, 3) 'D'(3, 3) 'E'(4, 3)
Approach 1 : Using brute force
Approach 2 : Using map of type char and int and store the frequency of characters
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Which SQL clause is used to specify the conditions in a query?