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# SUBARRAY - Editorial

Author: Devendra Agarwal
Tester: Anudeep Nekkanti
Editorialist: Amit Pandey

Medium.

# PREREQUISITES:

Dynamic programming and Data Structure(stack).

# PROBLEM:

You are given a character parenthesis ( having $[,],{,},<,>,(,)$) array and an integer array. You need to find the maximum sum sub-array in the integer array such that the corresponding sub-array in the character array has balanced parenthesis.

# QUICK EXPLANATION:

The given problem can be solved using a dynamic programming approach quite similar to maximum subarray problem. We need to take care of balanced parenthesis, which can be done using a classical approach.

# EXPLANATION:

First Problem:
How to solve maximum sum sub array problem using Kadane ALgorithm, which is a classical dynamic programming problem.

def max_subarray(A):
max_ending_here = max_so_far = 0
for x in A:
max_ending_here = max(0, max_ending_here + x)
max_so_far = max(max_so_far, max_ending_here)
return max_so_far

Second problem:
Given a character parenthesis array, check if it is balanced or not.

1) Declare a character stack $S$.
2) Now traverse the expression string expression.

• If the current character is a starting bracket then push it to stack.
• If the current character is a closing bracket, then pop from stack and if the popped character is the matching starting bracket then fine else parenthesis are not balanced.

3) After complete traversal, if there is some starting bracket left in stack then “not balanced”.

Original Problem:
Now back to original problem. Traverse the character array and if there is a closing brace at position $i$, determine the largest index($j < i$) such that $[j,i]$ is balanced. We can this in one pass of the character array using a stack. The approach is similar to the second problem given above.

for(int i=0;i<=N+1;i++)
hsh[i] = 0;     // This array will store j for each i.
stack<int> st;
st.push(0);
for (int i = 1; i <= n; i++) {
if (!st.empty()) {
// check if top of stack is opposite of current parenthesis.
if (closingBracket(s[i]) && s[st.top()] == opposite(s[i])) {
hsh[i] = st.top(); // We found j for the i.
st.pop();
} else {
st.push(i);
}
} else {
st.push(i);
}
}

When we are at index $i$, we may consider what we have solved problem upto index $i-1$, and we have created an array $DP[ \hspace{1 mm}]$. Where, $DP[j]$ stored the maximum sum ending at index $j$.

Dynamic Programming recurrence:

Let us call a sub-array valid, if its corresponding character array is balanced. Let dp[i] denotes maximum sum valid sub-array ending at position $i$. So recurrence for $DP$ will be as follows.

$DP[i] = max$ $(DP[i], dp[hsh[i] - 1] + sum(hsh[i], i))$

$hsh[i] \text{ is the largest j (j < i) such that segment} [j,i] \text{ is the balanced.}$

The given recurrence is using the simple fact if expressions $E_{1}$ and $E_{2}$ are balanced, expression $E_{1}E_{2}$ is balanced.

For finding out overall maximum sum sub-array we can iterate over each $i$ and take maximum of $DP[i]$.

# Solutions:

Setter's solution can be found here.
Tester's solution can be found here.

This question is marked "community wiki".

25911522
accept rate: 0%

19.8k350498541

I think you should note that \text{DP[i]} is initialized to 0 in the recurrence and the recurrence is only defined for \text{i} such that there is a balanced expression ending at \text{i}. Also, you never mention that \text{sum(hsh[i], i)} is the sum of the array from \text{hsh[i]} to \text{i}.

Anyway, thank you for the editorial! When I first saw this problem in the Cook-Off, I thought it was really hard, but now I see it wasn't that bad after all!

(20 Feb '15, 20:30)

 0 can anybody tell why this solution got tle solution answered 16 Feb '15, 00:30 2★anurag92 1●1●4 accept rate: 0% @Anurag92, You are declaring sum[100100] in each of the test case. While there can be 10^5 test cases. You should either keep it global or declare array of size N. (16 Feb '15, 00:33) thnx :) . should have declared outside (16 Feb '15, 00:42) anurag922★
 0 why largest index j in the hsh[] array? answered 16 Feb '15, 00:42 1●2 accept rate: 0% Sorry for the confusion, please check now. (16 Feb '15, 00:47) Actually i'm wondering why not for all j < i such that [j, i] is properly parenthesized? (16 Feb '15, 00:52) correct me if i'm mistaken anywhere :) (16 Feb '15, 00:52) according to the solution, you are trying to keep [j,i] segment length as minimum as possible (16 Feb '15, 00:54) Yes and segment [j,i] is balanced as well. (16 Feb '15, 13:01)
 0 shouldn't it be for all j < i such that [j, i] is properly parenthesized? answered 16 Feb '15, 00:49 1●2 accept rate: 0% Didn't get you, please elaborate. (16 Feb '15, 00:55) will there be a unique j for each i? Actually i tried making a balanced parenthesis but observed that j is always unique. if i tried making j lesser ( close to index 0 ), the segment [j,i] was getting imbalanced. I'll figure out my confusion. Thanks :) (16 Feb '15, 01:00) Yes, I am talkign about largest j, such that segment [j,i] is balanced. (16 Feb '15, 01:15)
 0 For each i, there will be only one j such that [j,i] will be balanced. I couldn't come up with a counter example, so I assume its true. answered 16 Feb '15, 01:05 1●2 accept rate: 0% Just try to realize it using the fact which is written in penultimate line. (16 Feb '15, 01:15) 1 Counterexample: ()(), For last parenthesis, there are two j's. (16 Feb '15, 04:02)
 0 can anybody tell why this solution got wa link text answered 16 Feb '15, 01:23 2★legar 1 accept rate: 0%
 0 I applied the same logic, but got tle. Can anybody suggest what's wrong in my solution answered 16 Feb '15, 08:42 1 accept rate: 0%
 0 j=i-1 or (flagvalue,say:-1) I don't think j can take any other value according to you.Am I correct? If we take j as the smallest j such that [j,i] is valid, then sum of elements from j,i should be the answer and that can be calculated in O(1) by keeping prefix sums array.Is this solution Correct? answered 20 Feb '15, 03:17 5★saurv4u 31●2 accept rate: 0% @saurv4u, take $j$ as the largest value such that $[j,i]$ is valid. In this way, you will be able to use dynamic programming approach. (20 Feb '15, 10:40)
 0 Used the exact same logic but got WA. If I find a closing bracket which does not match with the top most bracket in the stack, do I need to push it in the stack? I don't think it should make any difference. And can I know for which test cases did my code fail? http://www.codechef.com/viewsolution/6329799 answered 22 Feb '15, 02:23 46●2 accept rate: 10% I will read your code and will get back to you soon. :) (23 Feb '15, 10:57)
 0 Some test cases would be appreciated. I get 'Wrong Answer', but not sure for which cases. Not even sure if I'm reading the data incorrectly or just algorithm is wrong. Can somebody throw in a few 'tricky' test cases? answered 02 Feb '16, 12:18 0★madgod 1 accept rate: 0% Found the algo problem. Wanted to have [O(1) + stack] extra memory, naive :) If you want some test cases, I would recommend to write a function such as this to generate all combinations of input data of size N: Btw, can only use one type of brackets for this. char BA[] = { '{', '(', '[', '<', '>', ']', ')', '}' }; void GenSolve(int i) { if(i == N) { C[i] = '\0'; solve(); return; } if(i == 0) { for(int k = 0; k < N; ++k) A[k] = k + 1; } for(int b = 0; b < 8; ++b) { C[i] = BA[b]; GenSolve(i + 1); } } (03 Feb '16, 11:52) madgod0★
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question asked: 10 Feb '15, 01:52

question was seen: 5,882 times

last updated: 11 Feb '16, 19:13