Given a binary string s, return true if the longest contiguous segment of 1's is strictly longer than the longest contiguous segment of 0's in s, or return false otherwise.
s = "110100010 " the longest continuous segment of 1s has length 2, and the longest continuous segment of 0s has length 3.Note that if there are no 0's, then the longest continuous segment of 0's is considered to have a length 0. The same applies if there is no 1's.
Example 1:
Input: s = "1101 " Output: true Explanation: The longest contiguous segment of 1s has length 2: "1101 " The longest contiguous segment of 0s has length 1: "1101 " The segment of 1s is longer, so return true.
Example 2:
Input: s = "111000 " Output: false Explanation: The longest contiguous segment of 1s has length 3: "111000 " The longest contiguous segment of 0s has length 3: "111000 " The segment of 1s is not longer, so return false.
Example 3:
Input: s = "110100010 " Output: false Explanation: The longest contiguous segment of 1s has length 2: "110100010 " The longest contiguous segment of 0s has length 3: "110100010 " The segment of 1s is not longer, so return false.
Constraints:
1 <= s.length <= 100s[i] is either '0' or '1'.program main
! Solves the problem described in the comment block
implicit none
character(len=100) :: s
logical :: result
! Example 1
s = "1101 "
result = solve(s)
write (unit=*, fmt="(L1)") result
! Example 2
s = "111000 "
result = solve(s)
write (unit=*, fmt="(L1)") result
! Example 3
s = "110100010 "
result = solve(s)
write (unit=*, fmt="(L1)") result
contains
function solve(s) result(result)
! Returns true if the longest contiguous segment of 1s is strictly
! longer than the longest contiguous segment of 0s in s, or returns
! false otherwise
implicit none
character(len=*), intent(in) :: s
integer :: i, j, n, m
logical :: result
n = len(s)
m = 0
do i = 1, n
if (s(i:i) == '1') then
j = i
do while (j <= n .and. s(j:j) == '1')
j = j + 1
end do
if (j - i > m) then
m = j - i
end if
else if (s(i:i) == '0') then
j = i
do while (j <= n .and. s(j:j) == '0')
j = j + 1
end do
if (j - i > m) then
m = j - i
end if
end if
end do
result = m > 0 .and. m > n - m
end function solve
end program mainF F F
!-------------------------------------------------------------------------------
! Copyright (c) 2020, 2021, 2022, 2023, 2024, 2025, 2026, 2027, 2028, 2029, 2030,
! 2031, 2032, 2033, 2034, 2035, 2036, 2037, 2038, 2039, 2040, 2041, 2042, 2043,
! 2044, 2045, 2046, 2047, 2048, 2049, 2050, 2051, 2052, 2053, 2054, 2055, 2056,
! 2057, 2058, 2059, 2060, 2061, 2062, 2063, 2064, 2065, 2066, 2067, 2068, 2069,
! 2070, 2071, 2072, 2073, 2074, 2075, 2076, 2077, 2078, 2079, 2080, 2081, 2082,
! 2083, 2084, 2085, 2086, 2087, 2088, 2089, 2090, 2091, 2092, 2093, 2094, 2095,
! 2096, 2097, 2098, 2099, 2100, 2101, 2102, 2103, 2104, 2105, 2106, 2107, 2108,
! 2109, 2110, 2111, 2112, 2113, 2114, 2115, 2116, 2117, 2118, 2119, 2120, 2121,
! 2122, 2123, 2124, 2125, 2126, 2127, 2128, 2129, 2130, 2131, 2132, 2133, 2134,
! 2135, 2136, 2137, 2138, 2139, 2140, 2141, 2142, 2143, 2144, 2145, 2146, 2147,
! 2148, 2149, 2150, 2151, 2152, 2153, 2154, 2155, 2156, 2157, 2158, 2159, 2160,
! 2161, 2162, 2163, 2164, 2165, 2166, 2167, 2168, 2169, 2170, 2171, 2172, 2173,
! 2174, 2175, 2176, 2177, 2178, 2179, 2180, 2181, 2182, 2183,
/usr/bin/ld: /usr/lib/gcc/x86_64-linux-gnu/11/../../../x86_64-linux-gnu/Scrt1.o: in function `_start': (.text+0x1b): undefined reference to `main' collect2: error: ld returned 1 exit status
def checkZeroOnes(s: str) -> bool:
max_ones, max_zeros, current_ones, current_zeros = 0, 0, 0, 0
for c in s:
if c == '1':
current_ones += 1
current_zeros = 0
else:
current_zeros += 1
current_ones = 0
max_ones = max(max_ones, current_ones)
max_zeros = max(max_zeros, current_zeros)
return max_ones > max_zeros
We initialize 4 variables: max_ones, max_zeros, current_ones, and current_zeros. We then iterate through each character in the string. If we find 1, we increment the current_ones counter and set the current_zeros count to 0, and if we find 0, we increment the current_zeros counter and set the current_ones counter to 0. After each character, we update the maximum values of max_ones and max_zeros using the max function or Math.max function depending on the language. Finally, after iterating through the entire string, we compare the max_ones and max_zeros to determine if the longest continuous segment of 1s is strictly longer than that of 0s and return the result.
bool checkZeroOnes(string s) {
int max_ones = 0, max_zeros = 0, current_ones = 0, current_zeros = 0;
for (char c : s) {
if (c == '1') {
current_ones++;
current_zeros = 0;
} else {
current_zeros++;
current_ones = 0;
}
max_ones = max(max_ones, current_ones);
max_zeros = max(max_zeros, current_zeros);
}
return max_ones > max_zeros;
}
We initialize 4 variables: max_ones, max_zeros, current_ones, and current_zeros. We then iterate through each character in the string. If we find 1, we increment the current_ones counter and set the current_zeros count to 0, and if we find 0, we increment the current_zeros counter and set the current_ones counter to 0. After each character, we update the maximum values of max_ones and max_zeros using the max function or Math.max function depending on the language. Finally, after iterating through the entire string, we compare the max_ones and max_zeros to determine if the longest continuous segment of 1s is strictly longer than that of 0s and return the result.