Minimum Distance Between 1's

The 2 approaches which I had for the problem are mentioned below, basically the second approach gives a TC of O(n^2) and the first one gives a TC of O(n). The first one passes all the Test Cases whereas the second one fails 1 for Limit Exceeded.

# cook your dish here

T=int(input())

while(T):

    def get_dist(x1,y1):

        if(abs(x1-y1)%2!=0):

            return 1

        else:

            return 2

    n=int(input())

    st=input()

    s_len=len(st)

    m=float('inf')

    p1,p2,first=0,0,False

    for i in range(s_len):

        if(st[i]=='1' and first==True):

            p2=i

            if(get_dist(p1,p2)<m):

                m=get_dist(p1,p2)

                if(m==1):

                    break

        elif(st[i]=='1'):

            p1=i

            first=True

    print(m)

    T-=1

''' Second Approach '''

T=int(input("What's T? "))

while(T):

    n=int(input("What's n? "))

    st=input("What's st? ")

    indices=[]

    le=len(st)

    for i in range(le):

        if(st[i]=="1"):

            indices.append(i)

    le_ind=len(indices)

    # Now I have all the instances of 1 in the indices list 

    chk=False

    for i in range(le_ind-1):

        for j in range(i+1,le_ind):

            if(abs(indices[i]-indices[j])%2!=0):

                print(1)

                chk=True

                break

        if(chk==True):

            break

    if(chk==False):

        print(2)

    T-=1