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Program to check whether we can split a string into descending consecutive values in Python
Suppose we have a string s with only numbers. We have to check whether we can split s into two or more non-empty substrings such that the numerical values of those substrings are in non-increasing sequence and the difference between numerical values of every two adjacent substrings is 1. So for example, if the string is s = "0080079" we can split it into ["0080", "079"] with numerical values [80, 79]. And the values are in descending order and adjacent values differ by 1, so this way is valid. We have to check whether it is possible to split s as described above, or not.
So, if the input is like s = "080076", then the output will be True because we can split it like ["08", "007", "6"], so numeric values are [8,7,6].
To solve this, we will follow these steps −
Define a function dfs() . This will take s, pre, idx, n
if pre is not -1 and integer form of (substring of s[from index idx to end]) is same as pre -1, then
return True
for i in range 1 to n-idx-1, do
curs := substring of s[from index idx to idx+i-1]
cur := curs as number
if pre is same as -1, then
if dfs(s, cur, idx+i, n) is true, then
return True
otherwise,
if cur is same as pre - 1 and dfs(s, cur, idx+i, n) is true, then
return True
return False
From the main method, do the following
n := size of s
if n <= 1, then
return False
return dfs(s, -1, 0, n)
Example
Let us see the following implementation to get better understanding −
def dfs(s, pre, idx, n):
if pre != -1 and int(s[idx:]) == pre - 1:
return True
for i in range(1, n-idx):
curs = s[idx: idx+i]
cur = int(curs)
if pre == -1:
if dfs(s, cur, idx+i, n):
return True
else:
if cur == pre - 1 and dfs(s, cur, idx+i, n):
return True
return False
def solve(s):
n = len(s)
if n <= 1:
return False
return dfs(s, -1, 0, n)
s = "080076"
print(solve(s))
Input
"080076"
Output
True
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