unreasonably effective

link Two pointers Starting with the two pointers (right, left) = (31, 0), keep swapping towards the center. Time: , space: .

link Subsets of numbers For each subset we can xor the elements. If there are 3 numbers in the array, there are possible subsets. So, we use the bit patterns for the numbers to find elements of the subsets. Time: , space: . Subsets of bit positions Say we have two -bit numbers: and .…

link If we can mark the end of each word, we can decode. To mark the end of a word, we cannot use a character from the alphabet because we would not know if it is a marker or part of a word. Characters as numbers The ord() of a character is a non-negative integer.…

link We can create a map: {number -> count}. In the map, the number that has count = 1 is the one we want. Assuming numbers are 32-bit integers, there are distinct numbers. Time: , space: . Above, the count of numbers have a pattern: it is either multiple of 3 or 1. We can…

link If we xor all numbers, the number without a double will remain. Time: , space: .

link Horizontal flip of a row is like swapping the values at pairs of indices starting from the two ends going towards the center. So, if values of a pair of such indices are different, a flip followed by an invert behaves like two inverts in sequence — resulting in zero changes. Likewise, same values…

link To find the complement we can xor with all-ones mask. To create the all-ones mask, we first find the leftmost 1-bit in n. Say -th bit is the leftmost 1-bit. Then, will have all bits right of set to 1’s and that is our mask. Time: , space: . We can create the all-ones…

link Two passes Lookup t‘s characters in s to find which one is extra. Say len(s) = . Time: , space: . One pass Note, XOR of same numbers is 0. Also, XOR is commutative. So, if we take XOR of all characters (as number) in both strings, we shall be left with the extra…