字符串中重复之间的曼哈顿距离之和

给定一个由小写字符组成的大小为N的字符串S ，任务是找到每对(i, j)之间的曼哈顿距离之和，使得i≤j和S[j] = S[i] 。

例子：

Input: S = “ababa”
Output: 10
Explanation: The pairs having same characters are: (1, 3), (1, 5), (2, 4) and (3, 5). Therefore, the sum of Manhattan distance will be |3 – 1| + |5 – 1| + |4 – 2| + |5 – 3| = 10

Input: S = “abc”
Output: 0

朴素方法：最简单的方法是使用两个嵌套循环生成所有对(i, j)并检查每一对是否满足给定条件。如果发现是真的，将他们的距离添加到答案中。检查所有对后，打印答案。

时间复杂度： O(N ² )
辅助空间： O(1)

高效方法：该方法类似于查找所有点对之间的曼哈顿距离之和。请按照以下步骤解决问题：

• 将变量ans初始化为 0 以存储所需的结果。
• 为每个字符创建一个向量V以分别存储每个字符的位置。
• 遍历字符串S ，并将每个字符的位置附加到它们各自的向量V中。
• 现在，问题被简化为找到每个向量数组的每对点之间的曼哈顿距离之和。
• 使用变量i在[0, 25]范围内迭代
  • 将向量V[i]中存在的所有元素的总和存储在变量sum中。
  • 使用变量j遍历向量V[i]
    • 从sum中减去V[i][j]的值。
    • 将值sum – V[i][j] * (V[i].size() – 1 – j)添加到ans 。
• 打印ans的值作为结果。

Let the elements of the vector be x1, x2, x3, x4 which represents the indices of the same character.
This character will contribute value = |x2 – x1| + |x3 – x1| + |x4 – x1| + |x3 – x2| + | x4 – x2| + |x4 – x3|  

For a sorted array, this can also be written as (x2 + x3 + x4) – 3*x1 + (x3 + x4) – 2*x2 + (x4) – 1*x3. 

Now, the sum can also be expressed as ∑suffix[i + 1] – (n-i)*xi for i = 1 to n.
where suffix[i+1] is sum of elements from [i+1, n]. 

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the sum of the manhattan
// distances between same characters in string
void SumofDistances(string s)
{
    // Vector to store indices for each
    // unique character of the string
    vector v[26];
 
    // Append the position of each character
    // in their respective vectors
    for (int i = 0; i < s.size(); i++) {
        v[s[i] - 'a'].push_back(i);
    }
 
    // Store the required result
    int ans = 0;
 
    // Iterate over all the characters
    for (int i = 0; i < 26; i++) {
        int sum = 0;
 
        // Calculate sum of all elements
        // present in the vector
        for (int j = 0; j < v[i].size(); j++) {
            sum += v[i][j];
        }
 
        // Traverse the current vector
        for (int j = 0; j < v[i].size(); j++) {
 
            // Find suffix[i+1]
            sum -= v[i][j];
 
            // Adding distance of all pairs
            // whose first element is i and
            // second element belongs to [i+1, n]
            ans += (sum
                    - (v[i].size() - 1 - j) * (v[i][j]));
        }
    }
 
    // Print the result
    cout << ans;
}
 
// Driver Code
int main()
{
    // Given Input
    string s = "ababa";
 
    // Function Call
    SumofDistances(s);
 
    return 0;
}

Java

// Java program for the above approach
import java.lang.*;
import java.util.*;
 
class GFG{
     
// Function to find the sum of the manhattan
// distances between same characters in string
static void SumofDistances(String s)
{
     
    // Vector to store indices for each
    // unique character of the string
    ArrayList> v = new ArrayList<>();
     
    for(int i = 0; i < 26; i++)
        v.add(new ArrayList<>());
 
    // Append the position of each character
    // in their respective vectors
    for(int i = 0; i < s.length(); i++)
    {
        v.get(s.charAt(i) - 'a').add(i);
    }
 
    // Store the required result
    int ans = 0;
 
    // Iterate over all the characters
    for(int i = 0; i < 26; i++)
    {
        int sum = 0;
 
        // Calculate sum of all elements
        // present in the vector
        for(int j = 0; j < v.get(i).size(); j++)
        {
            sum += v.get(i).get(j);
        }
 
        // Traverse the current vector
        for(int j = 0; j < v.get(i).size(); j++)
        {
             
            // Find suffix[i+1]
            sum -= v.get(i).get(j);
 
            // Adding distance of all pairs
            // whose first element is i and
            // second element belongs to [i+1, n]
            ans += (sum - (v.get(i).size() - 1 - j) *
                          (v.get(i).get(j)));
        }
    }
 
    // Print the result
   System.out.println(ans);
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given Input
    String s = "ababa";
     
    // Function Call
    SumofDistances(s);
}
}
 
// This code is contributed by offbeat

Python3

# Python program for the above approach
 
 
#Function to find the sum of the manhattan
#distances between same characters in string
def SumofDistances(s):
   
    # Vector to store indices for each
    # unique character of the string
    v = [[] for i in range(26)]
 
    # Append the position of each character
    # in their respective vectors
    for i in range(len(s)):
        v[ord(s[i]) - ord('a')].append(i)
 
    # Store the required result
    ans = 0
 
    # Iterate over all the characters
    for i in range(26):
        sum = 0
 
        # Calculate sum of all elements
        # present in the vector
        for j in range(len(v[i])):
            sum += v[i][j]
 
        # Traverse the current vector
        for j in range(len(v[i])):
            # Find suffix[i+1]
            sum -= v[i][j]
 
            # Adding distance of all pairs
            # whose first element is i and
            # second element belongs to [i+1, n]
            ans += (sum - (len(v[i]) - 1 - j) * (v[i][j]))
 
    # Print the result
    print (ans)
 
# Driver Code
if __name__ == '__main__':
    # Given Input
    s = "ababa"
 
    # Function Call
    SumofDistances(s)
 
# This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find the sum of the manhattan
// distances between same characters in string
static void SumofDistances(string s)
{
     
    // Vector to store indices for each
    // unique character of the string
    List[] v = new List[26];
    for(int i = 0; i < 26; i++)
      v[i] = new List();
 
    // Append the position of each character
    // in their respective vectors
    for(int i = 0; i < s.Length; i++)
    {
        v[(int)s[i] - 97].Add(i);
    }
 
    // Store the required result
    int ans = 0;
 
    // Iterate over all the characters
    for(int i = 0; i < 26; i++)
    {
        int sum = 0;
 
        // Calculate sum of all elements
        // present in the vector
        for(int j = 0; j < v[i].Count; j++)
        {
            sum += v[i][j];
        }
 
        // Traverse the current vector
        for(int j = 0; j < v[i].Count; j++)
        {
             
            // Find suffix[i+1]
            sum -= v[i][j];
 
            // Adding distance of all pairs
            // whose first element is i and
            // second element belongs to [i+1, n]
            ans += (sum - (v[i].Count - 1 - j) *
                          (v[i][j]));
        }
    }
 
    // Print the result
    Console.Write(ans);
}
 
// Driver Code
public static void Main()
{
     
    // Given Input
    string s = "ababa";
     
    // Function Call
    SumofDistances(s);
}
}
 
// This code is contributed by ipg2016107

Javascript

// Javascript program for the above approach
 
// Function to find the sum of the manhattan
// distances between same characters in string
function SumofDistances(s)
{
    let v = [];
    for(let i = 0; i < 26; i++)
        v.push([]);
  
    // Append the position of each character
    // in their respective vectors
    for(let i = 0; i < s.length; i++)
    {
        v[s[i].charCodeAt(0) - 'a'.charCodeAt(0)].push(i);
    }
  
    // Store the required result
    let ans = 0;
  
    // Iterate over all the characters
    for(let i = 0; i < 26; i++)
    {
        let sum = 0;
  
        // Calculate sum of all elements
        // present in the vector
        for(let j = 0; j < v[i].length; j++)
        {
            sum += v[i][j];
        }
  
        // Traverse the current vector
        for(let j = 0; j < v[i].length; j++)
        {
              
            // Find suffix[i+1]
            sum -= v[i][j];
  
            // Adding distance of all pairs
            // whose first element is i and
            // second element belongs to [i+1, n]
            ans += (sum - (v[i].length - 1 - j) *
                          (v[i][j]));
        }
    }
  
    // Print the result
   document.write(ans);
}
 
// Driver code
// Given Input
let s = "ababa";
 
// Function Call
SumofDistances(s);
 
// This code is contributed by unknown2108

输出：

10

时间复杂度： O(N)
辅助空间： O(N)