IntwPrep
BAN USERLets see if we can re-phrase the question:
Divide the array into two EQUAL halves which has same AVERAGE
which means, the array is of even size, and the question can be rephrased as:
Divide the array into two EQUAL halves so that they have the same SUM
Algorithm:
1. compute the total sum of the array elements: lets say it is X
2. We need to find two subarrays of equal length which adds to X/2
3. Use subset sum algorithm to see if such subarrays exist
Probability of 1 slot having i elements P(1)=
(1/M)^i * ((M-1)/M)^N-i
Probability of 2 slots having i elements P(2)=
(1/M)^2i * ((M-1)/M)^N-2i
Probability of 3 slots having i elements P(3)=
(1/M)^3i * ((M-1)/M)^N-3i
...
and so on, so the probability of having A slot with i elements is: P(1)+P(2)+....+P(M)
but I am not sure of a formula to generalize this equation.
How about you have 26 trees, one for each alphabet,
Properties of the tree: It has a root, and n children
Root occurs before all its children.
In English tree for A has 25 children, B has 24 Children. Similarly keep building the trees and stop when the number of children in all the trees is 25*(25+1)/2=325. Now order the roots based on the number of children you have.
N - size of input
K - size of required set
M - size of memory
- build minheaps of size M, we will have K/M heaps.
- Within memory store the smallest value (say MIN), and a pointer to the minheap (say HP) with this smallest value
- When reading the input if you encounter a value less than MIN, then adjust HP to add the new element - Time complexity log M
- Search through K/M heaps to find new MIN and HP - time complexity log (K/M)
Time complexity will be N (log M + log (K/M))
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1) Find the number of ways the boards can be divided into K subsequent parts. Which is
2) Now for each combination, we get the minimum time by assigning the maximum sum subpart to the fastest painter. Using this condition compute the minimum time required for each division found in step 1
3) Keep track of the minimum value computed in step 2
Time complexity is:
- IntwPrep October 09, 2013