Maximum difference of zeros and ones in binary string
Given a binary string of 0s and 1s. The task is to find the length of substring which is having maximum difference of number of 0s and number of 1s (number of 0s – number of 1s). In case of all 1s print -1.
Input : S = "11000010001" Output : 6 From index 2 to index 9, there are 7 0s and 1 1s, so number of 0s - number of 1s is 6. Input : S = "1111" Output : -1
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.
The idea is to use Dynamic Programming to solve the problem.
Before that we will convert given binary string into integer array of value 1s and -1s, say arr. That can be easily done by traversing the given binary string and if ith index contain ‘0’ make -1 in corresponding position in array. Similarly, if ith index contain ‘1’, make 1 in the array.
Now, at each index i we need to make decision whether to take it or skip it. So, declare a 2D array of size n x 2, where n is the length of the given binary string, say dp[n].
dp[i] define the maximum value upto index i, when we skip the i-th index element. dp[i] define the maximum value upto index i after taking the i-th index element. Therefore, we can derive dp[i] as: dp[i] = max(dp[i+1], dp[i+1] + arr[i]) dp[i] = max(dp[i+1] + arr[i], 0)
For all ones we check this case explicitly.