Make a Palindrome by Inserting Characters - Google Top Interview Questions
Problem Statement :
Given a string s, return the minimum number of characters needed to be inserted so that the string becomes a palindrome. Constraints n ≤ 1,000 where n is the length of s Example 1 Input s = "radr" Output 1 Explanation We can insert "a" to get "radar"
Solution :
Solution in C++ :
int solve(string s) {
int n = s.length();
int dp[n + 1][n + 1];
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= n; j++) {
if (i == 0) {
dp[i][j] = j;
} else if (j == 0) {
dp[i][j] = i;
} else if (s[i - 1] == s[n - j]) {
dp[i][j] = dp[i - 1][j - 1];
} else {
dp[i][j] = 1 + min(dp[i][j - 1], dp[i - 1][j]);
}
}
}
return dp[n][n] / 2;
}
Solution in Java :
import java.util.*;
class Solution {
public int solve(String s) {
return s.length() - lps(s);
}
int lps(String st) {
char[] s = st.toCharArray();
int m = s.length;
int[][] dp = new int[m][m];
for (int i = 0; i < m; i++) dp[i][i] = 1;
for (int i = m - 1; i >= 0; i--) {
for (int j = i + 1; j < m; j++) {
if (s[i] == s[j]) {
dp[i][j] = dp[i + 1][j - 1] + 2;
} else {
dp[i][j] = Math.max(dp[i + 1][j], dp[i][j - 1]);
}
}
}
return dp[0][m - 1];
}
}
Solution in Python :
class Solution:
def solve(self, s):
"""
#1: Most popular approach: time and space complexity: O(n^2).
Let f[i][j] be the length of the longest palindromic subsequence in s[i:j+1]
f[i][j] = f[i+1][j-1] + 2 if s[i] == s[j]
= max(f[i+1][j], f[i][j-1]) otherwise
"""
n = len(s)
f = [[0] * n for _ in range(n)]
for size in range(1, n + 1):
for i in range(n - size + 1):
j = i + size - 1
if s[i] == s[j]:
if i + 1 > j - 1:
f[i][j] = size
else:
f[i][j] = f[i + 1][j - 1] + 2
else:
f[i][j] = max(f[i + 1][j], f[i][j - 1])
return n - f[0][n - 1]
"""
#2: A different transition function: time and space complexity: still O(n^2).
But this is a pathway to a better solution.
Let f[size][i] be the length of the longest palindromic subsequence in s[i:i+size]
f[size][i] = f[size-2][i+1] + 2 if s[i] == s[i+size-1]
= max(f[size-1][i+1], f[size-1][i]) otherwise
"""
n = len(s)
f = [[0] * n for _ in range(n + 1)]
for size in range(1, n + 1):
for i in range(n - size + 1):
j = i + size - 1
if s[i] == s[j]:
if size - 2 > 0:
f[size][i] = f[size - 2][i + 1] + 2
else:
f[size][i] = size
else:
f[size][i] = max(f[size - 1][i + 1], f[size - 1][i])
return n - f[n][0]
"""
#3: Based on #2, do rolling array to reduce space: time complexity: still O(n^2), space: O(n).
In #2, notice that f[size] only depends on f[size-1] and f[size-2], so instead of storing the whole
(n+1) arrays, we just need 3 arrays, f0, f1, f2. f0 is the current f[size], f1 is the f[size-1], and
f2 is the f[size-2].
"""
n = len(s)
f0 = [0] * n
f1 = [0] * n
f2 = [0] * n
for size in range(1, n + 1):
f0, f1, f2 = [0] * n, f0, f1
for i in range(n - size + 1):
j = i + size - 1
if s[i] == s[j]:
if size - 2 > 0:
f0[i] = f2[i + 1] + 2
else:
f0[i] = size
else:
f0[i] = max(f1[i + 1], f1[i])
return n - f0[0]
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