Longest common subsequence

def lcs_length(text1, text2):
    m = len(text2)
    prev = [0] * (m + 1)                       # row i - 1 of the table
    for ch in text1:
        cur = [0] * (m + 1)
        for j in range(1, m + 1):
            if ch == text2[j - 1]:
                cur[j] = prev[j - 1] + 1       # match: extend the diagonal
            else:
                cur[j] = max(prev[j], cur[j - 1])
        prev = cur
    return prev[m]

#include <algorithm>
#include <string>
#include <vector>

int solve(std::string text1, std::string text2) {
    const int m = text2.size();
    std::vector<int> prev(m + 1, 0), cur(m + 1, 0);
    for (char ch : text1) {
        for (int j = 1; j <= m; ++j)
            cur[j] = ch == text2[j - 1] ? prev[j - 1] + 1 : std::max(prev[j], cur[j - 1]);
        std::swap(prev, cur);
    }
    return prev[m];
}

class Solution {
    public int lcsLength(String text1, String text2) {
        int m = text2.length();
        int[] prev = new int[m + 1], cur = new int[m + 1];
        for (char ch : text1.toCharArray()) {
            for (int j = 1; j <= m; j++)
                cur[j] = ch == text2.charAt(j - 1) ? prev[j - 1] + 1 : Math.max(prev[j], cur[j - 1]);
            int[] t = prev; prev = cur; cur = t;
        }
        return prev[m];
    }
}

function lcsLength(text1, text2) {
  const m = text2.length;
  let prev = new Array(m + 1).fill(0);
  for (const ch of text1) {
    const cur = new Array(m + 1).fill(0);
    for (let j = 1; j <= m; j++) cur[j] = ch === text2[j - 1] ? prev[j - 1] + 1 : Math.max(prev[j], cur[j - 1]);
    prev = cur;
  }
  return prev[m];
}

fn solve(text1: &str, text2: &str) -> i64 {
    let b: Vec<char> = text2.chars().collect();
    let m = b.len();
    let mut prev = vec![0i64; m + 1];
    for ch in text1.chars() {
        let mut cur = vec![0i64; m + 1];
        for j in 1..=m {
            cur[j] = if ch == b[j - 1] { prev[j - 1] + 1 } else { prev[j].max(cur[j - 1]) };
        }
        prev = cur;
    }
    prev[m]
}

solve :: String -> String -> Int
solve a b = last (foldl row (replicate (length b + 1) 0) a)
  where
    -- one row of the table from the previous row, for the character ch of the first string
    row prev ch = scanl step 0 (zip3 b prev (tail prev))
      where step left (c, diag, up) = if c == ch then diag + 1 else max up left