DP on String
public boolean wordBreak(String s, List<String> wordDict) {
boolean[] dp = new boolean[s.length() + 1];
dp[0] = true;
for (int i = 1; i <= s.length(); i++) {
for (String w : wordDict) {
if (i - w.length() >= 0
&& s.substring(i - w.length(), i).equals(w)
&& dp[i - w.length()]) {
dp[i] = true;
}
}
}
return dp[s.length()];
}Q: the total number of ways to decode it.
dp[i] : the total number of ways to decode at i:
dp[0] = 1
dp[1] = 0 when "02" situation, dp[1] = 1 when other situation.
Transition: dp[i] = dp[i - 1] + dp[i - 2];
public int numDecodings(String s) {
int[] dp = new int[s.length() + 1];
dp[0] = 1;
dp[1] = s.charAt(0) == '0' ? 0 : 1;
for (int i = 2; i <= s.length(); i++) {
int one = Integer.valueOf(s.substring(i - 1, i));
int two = Integer.valueOf(s.substring(i - 2, i));
if (one >= 1 && one <= 9) {
dp[i] += dp[i - 1];
}
if (two <= 26 && two >= 10) {
dp[i] += dp[i - 2];
}
}
return dp[s.length()];
}dp[i][j] : longest common subsequence ends at i in text1, j in text2
if s.charAt(i) == t.charAt(j) : dp[i][j] = dp[i - 1][j - 1] + 1
else: dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1])
public int longestCommonSubsequence(String text1, String text2) {
int m = text1.length(), n = text2.length();
int[][] dp = new int[m + 1][n + 1];
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (text1.charAt(i - 1) == text2.charAt(j - 1)) {
dp[i][j] = dp[i - 1][j - 1] + 1;
} else {
dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
}
}
}
return dp[m][n];
}The question asks the minimum number of deletion to make word1 and word2 the same, this means to get Longest Common Subsequence of two strings and use the total lens - LCS:
public int minDistance(String word1, String word2) {
if (word1.equals(word2)) return 0;
int m = word1.length(), n = word2.length();
int[][] dp = new int[m + 1][n + 1];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (word1.charAt(i) == word2.charAt(j)) {
dp[i + 1][j + 1] = dp[i][j] + 1;
} else {
dp[i + 1][j + 1] = Math.max(dp[i][j + 1], dp[i + 1][j]);
}
}
}
return word1.length() + word2.length() - 2 * dp[m][n];
}dp[i][j] : maximum length of common subarray up to i and j
if (A[i] == B[j]) { dp[i][j] = dp[i - 1][j - 1] + 1; }
public int findLength(int[] A, int[] B) {
if (A == null || B == null || A.length == 0 || B.length == 0) return 0;
int m = A.length, n = B.length;
int[][] dp = new int[m + 1][n + 1];
int max = 0;
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (A[i - 1] == B[j - 1]) {
dp[i][j] = dp[i - 1][j - 1] + 1;
}
max = Math.max(max, dp[i][j]);
}
}
return max;
}when s.charAt(i) != t.charAt(j):
insert = dp[i][j] = dp[i - 1][j] + 1
delete = dp[i][j] = dp[i][j - 1] + 1
replace = dp[i][j] = dp[i - 1][j - 1] + 1
public int minDistance(String word1, String word2) {
if (word1 == null || word2 == null || word1.equals(word2)) return 0;
int m = word1.length(), n = word2.length();
int[][] dp = new int[m + 1][n + 1];
for (int i = 0; i <= m; i++) dp[i][0] = i;
for (int i = 0; i <= n; i++) dp[0][i] = i;
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
dp[i][j] = dp[i - 1][j - 1];
} else {
int insert = dp[i - 1][j] + 1;
int delete = dp[i][j - 1] + 1;
int replace = dp[i - 1][j - 1] + 1;
dp[i][j] = Math.min(insert, Math.min(delete, replace));
}
}
}
return dp[m][n];
}public boolean isMatch(String s, String p) {
int m = s.length(), n = p.length();
boolean[][] dp = new boolean[m + 1][n + 1];
dp[0][0] = true;
for (int i = 1; i <= m; i++) {
dp[i][0] = false;
}
for (int j = 1; j <= n; j++) {
if (p.charAt(j - 1) == '*') {
dp[0][j] = true;
} else {
break;
}
}
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (p.charAt(j - 1) != '*') {
dp[i][j] = dp[i - 1][j - 1] && (s.charAt(i - 1) == p.charAt(j - 1) || p.charAt(j - 1) == '?');
} else {
dp[i][j] = dp[i - 1][j] || dp[i][j - 1];
}
}
}
return dp[m][n];
}public boolean isMatch(String s, String p) {
int m = s.length(), n = p.length();
boolean[][] dp = new boolean[m + 1][n + 1];
dp[0][0] = true;
for (int i = 0; i < p.length(); i++) {
if (p.charAt(i) == '*' && dp[0][i-1]) {
dp[0][i+1] = true;
}
}
for (int i = 0; i < s.length(); i++) {
for (int j = 0; j < p.length(); j++) {
if (p.charAt(j) == s.charAt(i) || p.charAt(j) == '.') {
dp[i+1][j+1] = dp[i][j];
}
if (p.charAt(j) == '*') {
if (p.charAt(j-1) != s.charAt(i) && p.charAt(j-1) != '.') {
dp[i+1][j+1] = dp[i+1][j-1];
} else {
dp[i+1][j+1] = (dp[i+1][j] || dp[i][j+1] || dp[i+1][j-1]);
}
}
}
}
return dp[m][n];
}
dp[i][j] : the starting index of the substring where T has length i and S has length j.
if s.charAt(i) == t.charAt(j), dp[i][j] = dp[i - 1][j - 1]
else: dp[i][j] = dp[i - 1][j]
After find all starting indices, go through then and then length of the window is i - dp[i][n].
public String minWindow(String S, String T) {
int m = S.length(), n = T.length();
int[][] dp = new int[m + 1][n + 1];
for (int[] row : dp) Arrays.fill(row, -1);
for (int i = 0; i <= m; i++) {
dp[i][0] = i;
}
int start = -1, minLen = Integer.MAX_VALUE;
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= Math.min(i, n); j++) {
if (S.charAt(i - 1) == T.charAt(j - 1)) {
dp[i][j] = dp[i - 1][j - 1];
} else {
dp[i][j] = dp[i - 1][j];
}
}
if (dp[i][n] != -1) {
int len = i - dp[i][n];
if (len < minLen) {
minLen = len;
start = dp[i][n];
}
}
}
return start != -1 ? S.substring(start, start + minLen) : "";
}Palindrome
dp[i][j] = dp[i + 1][j - 1] + 2 when s.charAt(i) == s.charAt(j)
dp[i][j] = Math.max(dp[i + 1][j], dp[i][j - 1]) when s.charAt(i) != s.charAt(j)
public int longestPalindromeSubseq(String s) {
int n = s.length();
int[][] dp = new int[n][n];
for (int i = n - 1; i >= 0; i--) {
for (int j = i; j < n; j++) {
if (i == j) dp[i][j] = 1;
else {
if (s.charAt(i) == s.charAt(j)) {
dp[i][j] = dp[i + 1][j - 1] + 2;
} else {
dp[i][j] = Math.max(dp[i][j - 1], dp[i + 1][j]);
}
}
}
}
return dp[0][n - 1];
}public boolean isValidPalindrome(String s, int k) {
int n = s.length();
int[][] dp = new int[n][n];
for (int i = n - 1; i >= 0; i--) {
for (int j = i; j < n; j++) {
if (i == j) {
dp[i][j] = 1;
} else if (s.charAt(i) == s.charAt(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][n - 1] + k >= s.length();
}132. Palindrome Partitioning II
cut[R] is the minimum of cut[L - 1] + 1 (L <= R), if [L, R] is palindrome.
If [L, R] is palindrome, [L + 1, R - 1] is palindrome, and c[L] == c[R].
public int minCut(String s) {
int n = s.length();
boolean[][] dp = new boolean[n + 1][n + 1];
int[] min = new int[n];
for (int r = 0; r < n; r++) {
min[r] = r;
for (int l = 0; l <= r; l++) {
if (s.charAt(l) == s.charAt(r) && (r - l < 2 || dp[l + 1][r - 1])) {
dp[l][r] = true;
min[r] = l == 0 ? 0 : Math.min(min[r], min[l - 1] + 1);
}
}
}
return min[n - 1];
}Find how many palindromic subsequences (need not necessarily be distinct) can be formed in a given string. Note that the empty string is not considered as a palindrome.
Examples:
Input : str = "abcd"
Output : 4
Explanation :- palindromic subsequence are : "a" ,"b", "c" ,"d"
Input : str = "aab"
Output : 4
Explanation :- palindromic subsequence are :"a", "a", "b", "aa"
Input : str = "aaaa"
Output : 15
Initial Values : i= 0, j= n-1;
CountPS(i,j): Every single character of a string is a palindrome subsequence
if i == j: return 1 // palindrome of length 1
// If first and last characters are same, then we consider it as palindrome subsequence and checkfor the rest subsequence (i+1, j), (i, j-1)
Else if (str[i] == str[j)]: return countPS(i+1, j) + countPS(i, j-1) + 1;
public int countPS(String str) {
int N = str.length();
// create a 2D array to store the count
// of palindromic subsequence
int[][] cps = new int[N + 1][N + 1];
// palindromic subsequence of length 1
for (int i = 0; i < N; i++)
cps[i][i] = 1;
// check subsequence of length L is
// palindrome or not
for (int L = 2; L <= N; L++) {
for (int i = 0; i < N; i++) {
int k = L + i - 1;
if (k < N) {
if (str.charAt(i) == str.charAt(k)):
cps[i][k] = cps[i][k - 1]
+ cps[i + 1][k] + 1;
else:
cps[i][k] = cps[i][k - 1]
+ cps[i + 1][k]
- cps[i + 1][k - 1];
}
}
}
// return total palindromic subsequence
return cps[0][N - 1];
}
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