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pratyushk82010's blog

DP problem from NCR online coding round held on 3rd October
By pratyushk82010, history, 4 years ago, In English

Here is a DP problem on which I have struggled really bad but in vain. Hence, I would really appreciate you guys to help me out.

Problem : You are given a 2-D grid of n*m dimensions. You are requested to print the number of unique paths from top-left corner to the bottom-right corner such that the product of all the values in the path contains odd numbers of divisors. For each cell you can either move to right or down.

Input: n, m Output: Print a single integer denoting number of unique paths from top-left to bottom-right corner such that the product of all values in the path contains odd number of divisors. Print answer modulo 1e9 + 7.

Constraints: 1 <= n, m <= 100 value[i, j] <= 30

Example: Input: n = 3, m = 2

grid:  1 1
           3 1
           3 1 

      Output: 2

Explanation: Here are 2 unique paths: 1-> 3-> 3-> 1, Product = 9, Number of divisors of 9 are 1, 3, 9 which is odd. 1-> 1-> 1-> 1, Product = 1, Number of divisors of 1 is only 1 only, which is odd.

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Comments Comments (8)
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4 years ago, # |
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Sample Code
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    9 months ago, # ^ |
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    Can you please explain your approach?

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      9 months ago, # ^ |
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      Hey I've not done the problem myself but let me just talk about my idea real quick. If a number has odd number of divisors, the number would be a square number. So we can do something like bitmask since the value is small.

      Make a prime list for every prime $$$\leq 30$$$. Then let $$$dp_{i,j,k}$$$ to be the number of paths to reach cell $$$(i,j)$$$ with a bitmask $$$k$$$ where if a bit in $$$k$$$ is on, it means that the prime number's occurence would be odd. The transitions are obvious. Answer will be $$$dp_{n,m,0}$$$ because the every prime should appear even for a square number.

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        9 months ago, # ^ |
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        I am not able to understand the transitions. Can you please explain the transitions (in great detail if possible)?

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          9 months ago, # ^ |
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          $$$ a=\prod prime_i^{a_i} \\ b=\prod prime_i^{b_i} \\ a*b= \prod prime_i^{a_i+b_i} $$$

          Since we only need to know if its even/odd => ai+bi = ai ^ bi

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            9 months ago, # ^ |
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            Can you please explain the following lines of code, especially if(i) and if(j), like what they represent, why addition inside if(j) and not inside if(i), what does k^num represent?

            for(int k=0;k<1024;k++){ if(i){ dp[i][j][k]=dp[i-1][j][k^num]; } if(j){ dp[i][j][k]+=dp[i][j-1][k^num]; } }

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4 years ago, # |
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My accepted code
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