Sanket and Strings Problem

#1


Which case I am failing ??

#2

@ashwani225 Your approach is incorrect because you are assuming that all those min frequency elements will be less than k in the window that you are selecting to accomadate all max frequency elements.
Your code gives wrong answer for following test case.
1
abbaab

#3

Please tell the approach I am confused ??

#4

The logic for solving this problem is a simple one, we begin with 2 pointers left and right, we freeze left and increase right till it is possible to make string from left to right of one character once number of different character exceeds k, we move left pointer till it becomes less than k and then we freeze left pointer and move right and this process continues till right reaches n. We do this because this way we can find the maximum solution for each left.

#5

I didn’t get it Please explain and do tell that two pointers left and right both will be start from same location ??

#6

Take two-pointer l and r to mark the left and right index of the string under consideration.
starting from l=0,r=0,max=0,count=0.
repeat until r <n
3.1. increase the count whenever you find a different character(by different we mean if we are forming a string of an only, then b is different).
3.2. while count is greater than k,
3.2.1. decrement the count by one if the element at lth index is different.
3.2.1. increment l.
3.3. Compare max with count for maximum value.
3.4. increment r.

left act as left pointer and i as right pointer

This is the code ,please see https://ide.codingblocks.com/s/226002