CODEWITHSUNDEEP
pythonstring
Sagar Salunke
Sagar Salunke

Reputation: 11

Getting double of desired output

There is a binary string. We have to count number of substrings starting and ending with '1'. I have gone with this approach but not getting desired output Please tell me where does code fails.

Code-:

st="1001101"
c=0
for i in st:
    if i=='1':
        for j in st[st.index(i)+1:]:
            if j=='1':
                c+=1

print(c)

Upvotes: 0

Views: 43

Answers (2)

azro
azro

Reputation: 54168

The problem is with st.index(i) because as you have multiple 1 it always take the first

You may use enumerate to get both index and letter, for start letter

value = "1001101"
counter = 0
for idx, iletter in enumerate(value):
    if iletter == '1':
        for jletter in value[idx + 1:]:
            if jletter == '1':
                counter += 1
print(counter) # 6

Using a list comprehension, you can also compute the pairs of indexes that matches the requirement, then just take the length

pairs = [(i, j) for i in range(len(value)) 
                for j in range(i + 1, len(value)) 
                if value[i] == value[j] == "1"]

# [(0, 3), (0, 4), (0, 6), (3, 4), (3, 6), (4, 6)]
print(len(pairs)) #6

Upvotes: 2

COY
COY

Reputation: 704

This is in fact a math question...to know the answer, you only need to count number of 1's (call this c1), then the answer is nCr(n=c1, r=2)

st="1001101"
c1=0
for i in st:
    if i=='1': c1+=1

c=c1*(c1-1)//2

print(c)

BTW, the code in question fails because st.index(i) is always 1. You find the first occurrence of i='1' from st, but st starts with a '1' (or logically the first '1' from string). To correct it:

st="1001101"
c=0
st_len=len(st)
for i in range(0, st_len):
    if st[i]=='1':
        for j in range(i+1, st_len):
            if st[j]=='1':
                c+=1

print(c)

Upvotes: 1

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