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Program to find latest group of size M using Python
Suppose we have an array arr this is holding a permutation of numbers from 1 to n. If we have a binary string of size n and initially all its bits set to zero. Now at each step i (Indexing starts from 1 for both the binary string and arr) from 1 to n, the bit at position arr[i] is set to 1. We also have another value m, and we need to find the latest step at which there exists a group of ones of size m. Here a group of ones means a contiguous substring of 1s such that it cannot be extended in either direction. We have to find the latest step at which there exists a group of ones of length exactly m. If we cannot find any such group, then return -1.
So, if the input is like arr = [3,5,1,2,4] m = 3, then the output will be 4 because, initial binary string is "00000" then following steps −
"00100", groups: ["1"]
"00101", groups: ["1", "1"]
"10101", groups: ["1", "1", "1"]
"11101", groups: ["111", "1"]
"11111", groups: ["11111"]
Here the latest step at which there exists a group of size 3 is step 4.
To solve this, we will follow these steps −
n := size of arr
num := 0
ans := -1
l := an array of size n, and fill with 0
r := an array of size n, and fill with 0
for i in range 0 to n - 1, do
cur := 1
idx := arr[i] - 1
if r[idx] is same as m, then
num := num - 1
if l[idx] is same as m, then
num := num - 1
cur := cur + l[idx] + r[idx]
num := num + cur is same as m
if num > 0, then
ans := maximum of ans, i + 1
if idx - l[idx] > 0, then
r[idx - l[idx] - 1] := cur
if idx + r[idx] < n - 1, then
l[idx + r[idx] + 1] := cur
return ans
Let us see the following implementation to get better understanding −
Example
def solve(arr, m): n = len(arr) num = 0 ans = -1 l = [0] * n r = [0] * n for i in range(n): cur = 1 idx = arr[i] - 1 if r[idx] == m: num -= 1 if l[idx] == m: num -= 1 cur += l[idx] + r[idx] num += cur == m if num > 0: ans = max(ans, i + 1) if idx - l[idx] > 0: r[idx - l[idx] - 1] = cur if idx + r[idx] < n - 1: l[idx + r[idx] + 1] = cur return ans arr = [3,5,1,2,4] m = 3 print(solve(arr, m))
Input
[3,5,1,2,4], 3
Output
4
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