移動最小二乘

Problem statement:

問題陳述：

Given a string S, write a program to check if it is possible to construct the given string S by performing any of the below operations any number of times. In each step, you can:

給定字符串S ，編寫一個程序以檢查是否可以通過多次執行以下任何操作來構造給定的字符串S。 在每個步驟中，您可以：

Add any character at the end of the string.

在字符串末尾添加任何字符。

Append the existing string to the string itself.

將現有字符串追加到字符串本身。

The above steps can be applied any number of times. We need to print the minimum steps required to form the string.

以上步驟可以應用多次。 我們需要打印形成字符串所需的最少步驟。

Input:

輸入：

Input String

輸入字符串

Output:

輸出：

Minimum number of steps required to form that string.

形成該字符串所需的最少步驟數。

Constraints:

限制條件：

1 <= string length <= 10^5

Example:

例：

Input: aaaaaaaa Output: 4Explanation: Move 1: add 'a' to form "a" Move 2: add 'a' to form "aa" Move 3: append "aa" to form "aaaa" Move 4: append "aaaa" to form "aaaaaaaa"Input: ijhijhiOutput: 5Explanation: Move 1: add 'i' to form "i" Move 2: add 'j' to form "ij" Move 3: add 'h' to form "ijh" Move 4: append "ijh" to form "ijhijh" Move 5: add 'i' to form "ijhijhi"

Solution Approach:

解決方法：

So, we have two option

因此，我們有兩種選擇

Append single character

附加單個字符

Append the string itself

附加字符串本身

So, say the string is

所以說這個字符串是

x₁x₂ ... x_n

For any sub-problem of size i,

對于任何大小為i的子問題，

x₁x₂ ... x_i

Two things are possible,

有兩種可能

Appends x_i to sub-problem x₁x₂ ... x_(i-1)

將x _i附加到子問題x ₁ x ₂ ... x _(i-1)

Append x₁..x_j to x₁..x_j is x_(j+1)..x_i is similar to x₁..x_j

將x ₁ ..x _j追加到x ₁ ..x _j是x _{(j + 1)} ..x _i類似于x ₁ ..x _j

So, we can formulate the problem recursively,

因此，我們可以遞歸地提出問題，

f(i) = minimum steps to form x₁x₂ ... x_i f(i) = f(i-1)+1 f(i) = f(j)+1 if j=i/2 and the partitioned strings are same

Now find the minimum.

現在找到最小值。

The base value of f(i)=i obviously which is maximum value to form the string.

f(i)= i的基值顯然是構成字符串的最大值。

So, the above recursion can be transformed to DP.

因此，上述遞歸可以轉換為DP。

1) Declare int dp[n+1]; 2) Assign the base value. for i=0 to ndp[i]=i; 3) Fill the DP arrayfor i=2 to ndp[i]=minimum(dp[i-1]+1,dp[i]); // normal insertion, recursion first caseif i is evenif s₁s_(i/2) = s_(i/2+1)s_n // if appending string is possibledp[i]=minimum(dp[i],dp[i/2]+1);end ifend for4) return dp[n];DP[n] is the final result

C++ Implementation:

C ++實現：

#include <bits/stdc++.h> using namespace std;int minimumMoves(string s) {int n = s.length();int dp[n + 1];for (int i = 0; i <= n; i++)dp[i] = i;for (int i = 2; i <= n; i++) {dp[i] = std::min(dp[i - 1] + 1, dp[i]);if (i % 2 == 0) {int flag = 0;for (int j = 0; j < i / 2; j++) {if (s[j] != s[j + i / 2]) {flag = 1;break;}}if (flag == 0)dp[i] = std::min(dp[i], dp[i / 2] + 1);}}return dp[n]; }int main() {cout << "Enter input string\n";string s;cin >> s;cout << "Minimum steps to form te string: " << minimumMoves(s) << endl;return 0; }

Output:

輸出：

Enter input string incinclude Minimum steps to form te string: 8

翻譯自: https://www.includehelp.com/icp/minimal-moves-to-form-a-string.aspx

移動最小二乘

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