算法题：总和等于k的子数组数

2021年4月17日18:34:36 发表评论 1,085 次浏览

本文概述

给定一个未排序的整数数组, 找到总和等于给定数k的子数组的数量。

例子：

Input : arr[] = {10, 2, -2, -20, 10}, k = -10
Output : 3
Subarrays: arr[0...3], arr[1...4], arr[3..4]
have sum exactly equal to -10.

Input : arr[] = {9, 4, 20, 3, 10, 5}, k = 33
Output : 2
Subarrays : arr[0...2], arr[2...4] have sum
exactly equal to 33.

简单的解决方案–

一个简单的解决方案是遍历所有子数组并计算它们的总和。如果总和等于所需总和, 则增加子数组的数量。打印子数组的最终计数。 Java天真的解决方案的实现–

C++

//C++ program for
//the above approach
#include <bits/stdc++.h>
using namespace std;
int main()
{
   int arr[] = {10, 2, -2, -20, 10};
   int k = -10;
   int n = sizeof (arr) /sizeof (arr[0]);
   int res = 0;
 
   //Calculate all subarrays
   for ( int i = 0; i <n; i++)
   {
     int sum = 0;
     for ( int j = i; j <n; j++)
     {
       //Calculate required sum
       sum += arr[j];
 
       //Check if sum is equal to
       //required sum
       if (sum == k)
         res++;
     }
   }
   cout <<(res) <<endl;
}
 
//This code is contributed by Chitranayal

Java

import java.util.*;
class Solution {
     
     public static void main(String[] args)
     {
         int arr[] = { 10 , 2 , - 2 , - 20 , 10 };
         int k = - 10 ;
         int n = arr.length;
         int res = 0 ;
         
         //calculate all subarrays
         for ( int i = 0 ; i <n; i++) {
             
             int sum = 0 ;
             for ( int j = i; j <n; j++) {
                 
                 //calculate required sum
                 sum += arr[j];
                 
                 //check if sum is equal to
                 //required sum
                 if (sum == k)
                     res++;
             }
         }
         System.out.println(res);
     }
}

输出如下：

3

高效的解决方案–

一个有效的解决方案是遍历数组时, 将到目前为止的总和存储在求和中。另外, 请在地图中维护不同的求和值计数。如果在任何情况下, currsum的值等于所需的总和, 则子数组的个数增加1。 currsum的值超过currsum – sum的期望总和。如果从并购中删除此值, 则可以获得所需的总和。从图中找到先前发现的总和等于currsum-sum的子数组的数量。从当前子阵列中排除所有这些子阵列, 将得到具有所需总和的新子阵列。因此, 通过此类子数组的数量增加计数。请注意, 当currsum等于所需的总和时, 还请检查先前总和等于0的子数组的数量。从当前子数组中排除这些子数组会得到具有所需总和的新子数组。在这种情况下, 将计数增加总和为0的子数组的数量。

C++

//C++ program to find number of subarrays
//with sum exactly equal to k.
#include <bits/stdc++.h>
using namespace std;
 
//Function to find number of subarrays
//with sum exactly equal to k.
int findSubarraySum( int arr[], int n, int sum)
{
     //STL map to store number of subarrays
     //starting from index zero having
     //particular value of sum.
     unordered_map<int , int> prevSum;
 
     int res = 0;
 
     //Sum of elements so far.
     int currsum = 0;
 
     for ( int i = 0; i <n; i++) {
 
         //Add current element to sum so far.
         currsum += arr[i];
 
         //If currsum is equal to desired sum, //then a new subarray is found. So
         //increase count of subarrays.
         if (currsum == sum)
             res++;
 
         //currsum exceeds given sum by currsum
         // - sum. Find number of subarrays having
         //this sum and exclude those subarrays
         //from currsum by increasing count by
         //same amount.
         if (prevSum.find(currsum - sum) != prevSum.end())
             res += (prevSum[currsum - sum]);
 
         //Add currsum value to count of
         //different values of sum.
         prevSum[currsum]++;
     }
 
     return res;
}
 
int main()
{
     int arr[] = { 10, 2, -2, -20, 10 };
     int sum = -10;
     int n = sizeof (arr) /sizeof (arr[0]);
     cout <<findSubarraySum(arr, n, sum);
     return 0;
}

Java

//Java program to find number of subarrays
//with sum exactly equal to k.
import java.util.HashMap;
import java.util.Map;
 
public class GfG {
 
     //Function to find number of subarrays
     //with sum exactly equal to k.
     static int findSubarraySum( int arr[], int n, int sum)
     {
         //HashMap to store number of subarrays
         //starting from index zero having
         //particular value of sum.
         HashMap<Integer, Integer> prevSum = new HashMap<>();
 
         int res = 0 ;
 
         //Sum of elements so far.
         int currsum = 0 ;
 
         for ( int i = 0 ; i <n; i++) {
 
             //Add current element to sum so far.
             currsum += arr[i];
 
             //If currsum is equal to desired sum, //then a new subarray is found. So
             //increase count of subarrays.
             if (currsum == sum)
                 res++;
 
             //currsum exceeds given sum by currsum
             // - sum. Find number of subarrays having
             //this sum and exclude those subarrays
             //from currsum by increasing count by
             //same amount.
             if (prevSum.containsKey(currsum - sum))
                 res += prevSum.get(currsum - sum);
 
             //Add currsum value to count of
             //different values of sum.
             Integer count = prevSum.get(currsum);
             if (count == null )
                 prevSum.put(currsum, 1 );
             else
                 prevSum.put(currsum, count + 1 );
         }
 
         return res;
     }
 
     public static void main(String[] args)
     {
 
         int arr[] = { 10 , 2 , - 2 , - 20 , 10 };
         int sum = - 10 ;
         int n = arr.length;
         System.out.println(findSubarraySum(arr, n, sum));
     }
}
 
//This code is contributed by Rituraj Jain

Python3

# Python3 program to find the number of
# subarrays with sum exactly equal to k.
from collections import defaultdict
 
# Function to find number of subarrays 
# with sum exactly equal to k.
def findSubarraySum(arr, n, Sum ):
  
     # Dictionary to store number of subarrays
     # starting from index zero having 
     # particular value of sum.
     prevSum = defaultdict( lambda : 0 )
   
     res = 0
   
     # Sum of elements so far.
     currsum = 0
   
     for i in range ( 0 , n): 
   
         # Add current element to sum so far.
         currsum + = arr[i]
   
         # If currsum is equal to desired sum, # then a new subarray is found. So
         # increase count of subarrays.
         if currsum = = Sum : 
             res + = 1        
   
         # currsum exceeds given sum by currsum  - sum.
         # Find number of subarrays having 
         # this sum and exclude those subarrays
         # from currsum by increasing count by 
         # same amount.
         if (currsum - Sum ) in prevSum:
             res + = prevSum[currsum - Sum ]
           
   
         # Add currsum value to count of 
         # different values of sum.
         prevSum[currsum] + = 1
      
     return res
  
if __name__ = = "__main__" :
 
     arr =  [ 10 , 2 , - 2 , - 20 , 10 ] 
     Sum = - 10
     n = len (arr)
     print (findSubarraySum(arr, n, Sum ))
     
# This code is contributed by Rituraj Jain

C#

//C# program to find number of subarrays
//with sum exactly equal to k.
using System;
using System.Collections.Generic;
 
class GFG {
     //Function to find number of subarrays
     //with sum exactly equal to k.
     public static int findSubarraySum( int [] arr, int n, int sum)
     {
 
         //HashMap to store number of subarrays
         //starting from index zero having
         //particular value of sum.
         Dictionary<int , int> prevSum = new Dictionary<int , int>();
 
         int res = 0;
 
         //Sum of elements so far
         int currsum = 0;
 
         for ( int i = 0; i <n; i++) {
 
             //Add current element to sum so far.
             currsum += arr[i];
 
             //If currsum is equal to desired sum, //then a new subarray is found. So
             //increase count of subarrays.
             if (currsum == sum)
                 res++;
 
             //currsum exceeds given sum by currsum
             //- sum. Find number of subarrays having
             //this sum and exclude those subarrays
             //from currsum by increasing count by
             //same amount.
             if (prevSum.ContainsKey(currsum - sum))
                 res += prevSum[currsum - sum];
 
             //Add currsum value to count of
             //different values of sum.
             if (!prevSum.ContainsKey(currsum))
                 prevSum.Add(currsum, 1);
             else {
                 int count = prevSum[currsum];
                 prevSum[currsum] = count + 1;
             }
         }
         return res;
     }
 
     //Driver Code
     public static void Main()
     {
         int [] arr = { 10, 2, -2, -20, 10 };
         int sum = -10;
         int n = arr.Length;
         Console.Write(findSubarraySum(arr, n, sum));
     }
}
 
//This code is contributed by
//sanjeev2552

输出如下：

3

时间复杂度：O(logn)

辅助空间：O(n)

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