1.0 Introduction

Introduction

In many problems dealing with an array (or a LinkedList), we are asked to find or calculate something among all the contiguous subarrays (or sublists) of a given size. For example, take a look at this problem:

Given an array, find the average of all contiguous subarrays of size ‘K’ in it.

Let’s understand this problem with a real input: Array: [1, 3, 2, 6, -1, 4, 1, 8, 2], K=5 Here, we are asked to find the average of all contiguous subarrays of size ‘5’ in the given array. Let’s solve this:

For the first 5 numbers (subarray from index 0-4), the average is: (1+3+2+6-1)/5 => 2.2(1+3+2+6−1)/5=>2.2 The average of next 5 numbers (subarray from index 1-5) is: (3+2+6-1+4)/5 => 2.8(3+2+6−1+4)/5=>2.8 For the next 5 numbers (subarray from index 2-6), the average is: (2+6-1+4+1)/5 => 2.4(2+6−1+4+1)/5=>2.4

Here is the final output containing the averages of all contiguous subarrays of size 5: Output: [2.2, 2.8, 2.4, 3.6, 2.8]

A brute-force algorithm will calculate the sum of every 5-element contiguous subarray of the given array and divide the sum by ‘5’ to find the average. This is what the algorithm will look like:

import java.util.Arrays;

class AverageOfSubarrayOfSizeK {
  public static double[] findAverages(int K, int[] arr) {
    double[] result = new double[arr.length - K + 1];
    for (int i = 0; i <= arr.length - K; i++) {
      // find sum of next 'K' elements
      double sum = 0;
      for (int j = i; j < i + K; j++)
        sum += arr[j];
      result[i] = sum / K; // calculate average
    }

    return result;
  }

  public static void main(String[] args) {
    double[] result = AverageOfSubarrayOfSizeK.findAverages(5, new int[] { 1, 3, 2, 6, -1, 4, 1, 8, 2 });
    System.out.println("Averages of subarrays of size K: " + Arrays.toString(result));
  }
}

Time complexity: Since for every element of the input array, we are calculating the sum of its next ‘K’ elements, the time complexity of the above algorithm will be O(N*K) where ‘N’ is the number of elements in the input array.

Can we find a better solution? Do you see any inefficiency in the above approach?

The inefficiency is that for any two consecutive subarrays of size ‘5’, the overlapping part (which will contain four elements) will be evaluated twice. For example, take the above-mentioned input:

image

As you can see, there are four overlapping elements between the subarray (indexed from 0-4) and the subarray (indexed from 1-5). Can we somehow reuse the sum we have calculated for the overlapping elements?

The efficient way to solve this problem would be to visualize each contiguous subarray as a sliding window of ‘5’ elements. This means that we will slide the window by one element when we move on to the next subarray. To reuse the sum from the previous subarray, we will subtract the element going out of the window and add the element now being included in the sliding window. This will save us from going through the whole subarray to find the sum and, as a result, the algorithm complexity will reduce to O(N).

image

Here is the algorithm for the Sliding Window approach:

import java.util.Arrays;

class AverageOfSubarrayOfSizeK {
  public static double[] findAverages(int K, int[] arr) {
    double[] result = new double[arr.length - K + 1];
    double windowSum = 0;
    int windowStart = 0;
    for (int windowEnd = 0; windowEnd < arr.length; windowEnd++) {
      windowSum += arr[windowEnd]; // add the next element
      // slide the window, we don't need to slide if we've not hit the required window size of 'k'
      if (windowEnd >= K - 1) {
        result[windowStart] = windowSum / K; // calculate the average
        windowSum -= arr[windowStart]; // subtract the element going out
        windowStart++; // slide the window ahead
      }
    }

    return result;
  }

  public static void main(String[] args) {
    double[] result = AverageOfSubarrayOfSizeK.findAverages(5, new int[] { 1, 3, 2, 6, -1, 4, 1, 8, 2 });
    System.out.println("Averages of subarrays of size K: " + Arrays.toString(result));
  }
}

Template to solve some sliding window problem:

public class Solution {
    public List<Integer> slidingWindowTemplateByHarryChaoyangHe(String s, String t) {
        //init a collection or int value to save the result according the question.
        List<Integer> result = new LinkedList<>();
        if(t.length()> s.length()) return result;
        
        //create a hashmap to save the Characters of the target substring.
        //(K, V) = (Character, Frequence of the Characters)
        Map<Character, Integer> map = new HashMap<>();
        for(char c : t.toCharArray()){
            map.put(c, map.getOrDefault(c, 0) + 1);
        }
        //maintain a counter to check whether match the target string.
        int counter = map.size();//must be the map size, NOT the string size because the char may be duplicate.
        
        //Two Pointers: begin - left pointer of the window; end - right pointer of the window
        int begin = 0, end = 0;
        
        //the length of the substring which match the target string.
        int len = Integer.MAX_VALUE; 
        
        //loop at the begining of the source string
        while(end < s.length()){
            
            char c = s.charAt(end);//get a character
            
            if( map.containsKey(c) ){
                map.put(c, map.get(c)-1);// plus or minus one
                if(map.get(c) == 0) counter--;//modify the counter according the requirement(different condition).
            }
            end++;
            
            //increase begin pointer to make it invalid/valid again
            while(counter == 0 /* counter condition. different question may have different condition */){
                
                char tempc = s.charAt(begin);//***be careful here: choose the char at begin pointer, NOT the end pointer
                if(map.containsKey(tempc)){
                    map.put(tempc, map.get(tempc) + 1);//plus or minus one
                    if(map.get(tempc) > 0) counter++;//modify the counter according the requirement(different condition).
                }
                
                /* save / update(min/max) the result if find a target*/
                // result collections or result int value
                
                begin++;
            }
        }
        return result;
    }
}