How to minimize o(n^2) complexity in this problem test case1&2 t.t.l

#1

#include
#include
using namespace std;
int main() {
int k;
cin>>k;
string str;
cin>>str;
int r=0;
int q=0;
int n=str.length();
int max=0;
for(int i=0;i<n;i++){
string s=" ";
int q=0;
int r=0;
for(int j=i;j<n;j++){
s=s+str[j];
if(str[j]==‘a’){
r=r+1;
}
else{
q=q+1;
}

            if(q>r){
                if(r<=k){
                    if(max<r+q){
                        max=r+q;
                    }
                }
            }
            else{
                if(q<=k){
                    if(max<q+r){
                        max=q+r;
                    }
                }
            }
        }
        
    
}
cout<<max;
return 0;

}

#2

Hello @dineshjani,

To reduce the time complexity,
You can implement the following approach:
Perform following for both the characters a and b individually,

1.Take two-pointer l and r to mark the left and right index of the string under consideration.
2. starting from l=0,r=0,max=0,count=0.
3. 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.

Let’s understand this with an example:
Example:
1
abba
Output:
3
Explanation:

l=0: variable to mark the left index and initialized to 0
r=0: variable to mark the right index and initialized to 0
m=0: to store the perfectness and initialised to zero
t=k: to keep track of the number of replacements we can perform at any instance of time.

  1. First, we will check for the string of character ‘a’:

    character at a[r] is ‘a’, increment r
    r:1
    the character at a[r] is not ‘a’ and t>0, increment r and decrement t as we can replace one ‘b’ to ‘a’
    t:0 r:2
    the character at a[r] is not ‘a’ and t==0. So, we cannot replace. Thus replace m by max. of m and r-l.
    Repeat, while t==0:
    A. if a[l]!=‘a’: increment t (because we must have done one replacement for this before and now as we are not considering this character so we can use this replacement somewhere else.)
    B. increment l for each iteration of the loop.
    m:2 t:0 l:1 t:1 l:2
    the character at a[r] is not ‘a’ and t>0, increment r and decrement t as we can replace one ‘b’ to ‘a’
    t:0 r:3
    character at a[r] is ‘a’, increment r
    r:4
    replace m by max. of m and r-l.
    m:2

l=0: variable to mark the left index and initialized to 0
r=0: variable to mark the right index and initialized to 0
m=0: to store the perfectness and initialised to zero
t=k: to keep track of the number of replacements we can perform at any instance of time.

  1. Now, we will check for the string of character ‘b’:

    the character at a[r] is not ‘b’ and t>0, increment r and decrement t as we can replace one ‘a’ to ‘b’
    t:0 r:1
    character at a[r] is ‘b’, increment r
    r:2
    character at a[r] is ‘b’, increment r
    r:3
    the character at a[r] is not ‘b’ and t==0. So, we cannot replace. Thus replace m by max. of m and r-l.
    Repeat, while t==0:
    A. if a[l]!=‘b’: increment t (because we must have done one replacement for this before and now as we are not considering this character so we can use this replacement somewhere else.)
    B. increment l for each iteration of the loop.
    m:3 t:1 l:1
    the character at a[r] is not ‘b’ and t>0, increment r and decrement t as we can replace one ‘a’ to ‘b’
    t:0 r:4
    replace m by max. of m and r-l.
    m:3
    Hence, the output is m:
    3

Let me know if you want me to correct your code.
Hope, this would help.
Give a like if you are satisfied.

#3

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