# Check if characters of the given string can be rearranged to form a palindrome. # Counter is faster for long strings and non-Counter is faster for short strings.
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if(s1[i-1] == s2[j-1]) return 1 + f(i-1,j-1, s1, s2); return max(f(i-1,j, s1, s2), f(i,j-1, s1, s2)); if(dp[i][j] != 0) return dp[i][j]; if(s1[i-1] == s2[j-1]) return ...