sayantanc85
BAN USERBecause q/s is about efficient DS. Trie is most efficient in handling dictionary than a hash table
- sayantanc85 September 12, 2017Psuedocode:
int index = 0;
function(A, min, max):
node = null;
if(A[index] between min and mx)
node= A[index]
index++
node.left = function(A, min,A[index])
node.right = function(A, A[index], max)
return node;
I think this should work.
Go over all the digits one by one in the binary number given. As you scan the number, keep track of the count of even place 1's and odd place 1's.
Finally remainder will be odd count - even count. If -ve, remained is 3+ (-ve remainder)
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I would suggest a Trie instead of a hash table to hold the word to value mapping and use it to decode for evaluation of an expression.
- sayantanc85 September 12, 2017