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LeetCode link: 583. Delete Operation for Two Strings, difficulty: Medium.
Given two strings word1
and word2
, return the minimum number of steps required to make word1
and word2
the same.
In one step, you can delete exactly one character in either string.
Example 1:
Input: word1 = "sea", word2 = "eat"
Output: 2
Explanation:
You need one step to make "sea" to "ea" and another step to make "eat" to "ea".
Example 2:
Input: word1 = "leetcode", word2 = "etco"
Output: 4
Constraints:
1 <= word1.length, word2.length <= 500
word1
andword2
consist of only lowercase English letters.
Intuition
It is a question of comparing two strings. After doing similar questions many times, we will develop an intuition to use dynamic programming with two-dimensional arrays
.
Pattern of "Dynamic Programming"
"Dynamic programming" is divided into five steps
- Determine the meaning of each value of the array
dp
. - Initialize the value of the array
dp
. - Fill in the
dp
grid data "in order" according to an example. - Based on the
dp
grid data, derive the "recursive formula". - Write a program and print the
dp
array. If it is not as expected, adjust it.
Detailed description of these five steps
- Determine the meaning of each value of the array
dp
.- First determine whether
dp
is a one-dimensional array or a two-dimensional array. Aone-dimensional rolling array
means that the values of the array are overwritten at each iteration. Most of the time, usingone-dimensional rolling array
instead oftwo-dimensional array
can simplify the code; but for some problems, such as operating "two swappable arrays", for the sake of ease of understanding, it is better to usetwo-dimensional array
. - Try to use the meaning of the
return value
required by the problem as the meaning ofdp[i]
(one-dimensional) ordp[i][j]
(two-dimensional). It works about 60% of the time. If it doesn't work, try other meanings. - Try to save more information in the design. Repeated information only needs to be saved once in a
dp[i]
. - Use simplified meanings. If the problem can be solved with
boolean value
, don't usenumeric value
.
- First determine whether
- Initialize the value of the array
dp
. The value ofdp
involves two levels:- The length of
dp
. Usually:condition array length plus 1
orcondition array length
. - The value of
dp[i]
ordp[i][j]
.dp[0]
ordp[0][0]
sometimes requires special treatment.
- The length of
- Fill in the
dp
grid data "in order" according to an example.- The "recursive formula" is the core of the "dynamic programming" algorithm. But the "recursive formula" is obscure. If you want to get it, you need to make a table and use data to inspire yourself.
- If the original example is not good enough, you need to redesign one yourself.
- According to the example, fill in the
dp
grid data "in order", which is very important because it determines the traversal order of the code. - Most of the time, from left to right, from top to bottom. But sometimes it is necessary to traverse from right to left, from bottom to top, from the middle to the right (or left), such as the "palindrome" problems. Sometimes, it is necessary to traverse a line twice, first forward and then backward.
- When the order is determined correctly, the starting point is determined. Starting from the starting point, fill in the
dp
grid data "in order". This order is also the order in which the program processes. - In this process, you will get inspiration to write a "recursive formula". If you can already derive the formula, you do not need to complete the grid.
- Based on the
dp
grid data, derive the "recursive formula".- There are three special positions to pay attention to:
dp[i - 1][j - 1]
,dp[i - 1][j]
anddp[i][j - 1]
, the currentdp[i][j]
often depends on them. - When operating "two swappable arrays", due to symmetry, we may need to use
dp[i - 1][j]
anddp[i][j - 1]
at the same time.
- There are three special positions to pay attention to:
- Write a program and print the
dp
array. If it is not as expected, adjust it.- Focus on analyzing those values that are not as expected.
After reading the above, do you feel that "dynamic programming" is not that difficult? Try to solve this problem. 🤗
Steps
Common steps in dynamic programming
These five steps are a pattern for solving dynamic programming
problems.
- Determine the meaning of the
dp[i][j]
- Since there are two strings, we can use two-dimensional arrays as the default option.
- At first, try to use the problem's
return
value as the value ofdp[i][j]
to determine the meaning ofdp[i][j]
. If it doesn't work, try another way. dp[i][j]
represents the minimum number of steps required to makeword1
's firsti
letters andword2
's firstj
letters the same.dp[i][j]
is an integer.
Determine the
dp
array's initial valueUse an example:
After initialization, the 'dp' array would be: # e a t # 0 1 2 3 # dp[0] # s 1 0 0 0 # e 2 0 0 0 # a 3 0 0 0
dp[0][j] = j
, becausedp[0]
represents the empty string, and the number of steps is just the number of chars to be deleted.dp[i][0] = i
, the reason is the same as the previous line, just viewed in vertical direction.
Determine the
dp
array's recurrence formulaTry to complete the grid. In the process, you will get inspiration to derive the formula.
1. word1 = "s", word2 = "eat" # e a t # 0 1 2 3 # s 1 2 3 4 # dp[1]
2. word1 = "se", word2 = "eat" # e a t # 0 1 2 3 # s 1 2 3 4 # e 2 1 2 3
3. word1 = "sea", word2 = "eat" # e a t # 0 1 2 3 # s 1 2 3 4 # e 2 1 2 3 # a 3 2 1 2
When analyzing the sample
dp
grid, remember there are three important points which you should pay special attention to:dp[i - 1][j - 1]
,dp[i - 1][j]
anddp[i][j - 1]
. The currentdp[i][j]
often depends on them.If the question is also true in reverse (swap
word1
andword2
), and we need to usedp[i - 1][j]
ordp[i][j - 1]
, then we probably need to use both of them.We can derive the
Recurrence Formula
:if word1[i - 1] == word2[j - 1] dp[i][j] = dp[i - 1][j - 1] else dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + 1
Determine the
dp
array's traversal orderdp[i][j]
depends ondp[i - 1][j - 1]
,dp[i - 1][j]
anddp[i][j - 1]
, so we should traverse thedp
array from top to bottom, then from left to right.
Check the
dp
array's value- Print the
dp
to see if it is as expected.
- Print the
Complexity
Time complexity
O(N * M)
Space complexity
O(N * M)
Java #
class Solution {
public int minDistance(String word1, String word2) {
var dp = new int[word1.length() + 1][word2.length() + 1];
for (var i = 0; i < dp.length; i++) {
dp[i][0] = i;
}
for (var j = 0; j < dp[0].length; j++) {
dp[0][j] = j;
}
for (var i = 1; i < dp.length; i++) {
for (var j = 1; j < dp[0].length; j++) {
if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
dp[i][j] = dp[i - 1][j - 1];
} else {
dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + 1;
}
}
}
return dp[dp.length - 1][dp[0].length - 1];
}
}
C# #
public class Solution
{
public int MinDistance(string word1, string word2)
{
var dp = new int[word1.Length + 1, word2.Length + 1];
for (var i = 0; i < dp.GetLength(0); i++)
dp[i, 0] = i;
for (var j = 0; j < dp.GetLength(1); j++)
dp[0, j] = j;
for (var i = 1; i < dp.GetLength(0); i++)
{
for (var j = 1; j < dp.GetLength(1); j++)
{
if (word1[i - 1] == word2[j - 1])
{
dp[i, j] = dp[i - 1, j - 1];
}
else
{
dp[i, j] = Math.Min(dp[i - 1, j], dp[i, j - 1]) + 1;
}
}
}
return dp[dp.GetUpperBound(0), dp.GetUpperBound(1)];
}
}
Python #
class Solution:
def minDistance(self, word1: str, word2: str) -> int:
dp = [[0] * (len(word2) + 1) for _ in range(len(word1) + 1)]
for i in range(len(dp)):
dp[i][0] = i
for j in range(len(dp[0])):
dp[0][j] = j
for i in range(1, len(dp)):
for j in range(1, len(dp[0])):
if word1[i - 1] == word2[j - 1]:
dp[i][j] = dp[i - 1][j - 1]
else:
dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + 1
return dp[-1][-1]
C++ #
class Solution {
public:
int minDistance(string word1, string word2) {
vector<vector<int>> dp(word1.size() + 1, vector<int>(word2.size() + 1));
for (auto i = 0; i < dp.size(); i++) {
dp[i][0] = i;
}
for (auto j = 0; j < dp[0].size(); j++) {
dp[0][j] = j;
}
for (auto i = 1; i < dp.size(); i++) {
for (auto j = 1; j < dp[0].size(); j++) {
if (word1[i - 1] == word2[j - 1]) {
dp[i][j] = dp[i - 1][j - 1];
} else {
dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + 1;
}
}
}
return dp[dp.size() - 1][dp[0].size() - 1];
}
};
JavaScript #
var minDistance = function (word1, word2) {
const dp = Array(word1.length + 1).fill().map(
() => Array(word2.length + 1).fill(0)
)
dp.forEach((_, i) => { dp[i][0] = i })
dp[0].forEach((_, j) => { dp[0][j] = j })
for (let i = 1; i < dp.length; i++) {
for (let j = 1; j < dp[0].length; j++) {
if (word1[i - 1] == word2[j - 1]) {
dp[i][j] = dp[i - 1][j - 1]
} else {
dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + 1
}
}
}
return dp.at(-1).at(-1)
};
Go #
func minDistance(word1 string, word2 string) int {
dp := make([][]int, len(word1) + 1)
for i := range dp {
dp[i] = make([]int, len(word2) + 1)
dp[i][0] = i
}
for j := range dp[0] {
dp[0][j] = j
}
for i := 1; i < len(dp); i++ {
for j := 1; j < len(dp[0]); j++ {
if word1[i - 1] == word2[j - 1] {
dp[i][j] = dp[i - 1][j - 1]
} else {
dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + 1
}
}
}
return dp[len(dp) - 1][len(dp[0]) - 1]
}
Ruby #
def min_distance(word1, word2)
dp = Array.new(word1.size + 1) do
Array.new(word2.size + 1, 0)
end
dp.each_with_index do |_, i|
dp[i][0] = i
end
dp[0].each_with_index do |_, j|
dp[0][j] = j
end
(1...dp.size).each do |i|
(1...dp[0].size).each do |j|
dp[i][j] =
if word1[i - 1] == word2[j - 1]
dp[i - 1][j - 1]
else
[ dp[i - 1][j], dp[i][j - 1] ].min + 1
end
end
end
dp[-1][-1]
end