How to Sum Consecutives in Java

2 min read 400 words

The challenge

You are given a list/array which contains only integers (positive and negative). Your job is to sum only the numbers that are the same and consecutive. The result should be one list.

Extra credit if you solve it in one line. You can assume there is never an empty list/array and there will always be an integer.

Same meaning: 1 == 1

1 != -1

Examples:

[1,4,4,4,0,4,3,3,1] # should return [1,12,0,4,6,1]

"""So as you can see sum of consecutives 1 is 1 
sum of 3 consecutives 4 is 12 
sum of 0... and sum of 2 
consecutives 3 is 6 ..."""

[1,1,7,7,3] # should return [2,14,3]
[-5,-5,7,7,12,0] # should return [-10,14,12,0]

The solution in Java code

Option 1:

import java.util.*;
public class Consecutives {
    public static List<Integer> sumConsecutives(List<Integer> s) {
      int previous = Integer.MAX_VALUE;
      LinkedList<Integer> l = new LinkedList<>();
      for (Integer v: s){
         l.add(v == previous? l.pollLast() + v : v); 
         previous = v;
      }
      return l; 
    }
}

Option 2:

import java.util.ArrayList;
import java.util.List;
public class Consecutives {
    public static List<Integer> sumConsecutives(List<Integer> s) {
        List<Integer> accumulator = new ArrayList<>();
        for (int i = 0, sum = 0; i < s.size(); i++) {
            sum += s.get(i);
            if (i == s.size() - 1 || s.get(i) != s.get(i + 1)) {
                accumulator.add(sum);
                sum = 0;
            }
        }
        return accumulator;
    }
}

Option 3:

import java.util.*;
public class Consecutives {
    public static List<Integer> sumConsecutives(List<Integer> list) {
        List<Integer> result = new ArrayList<>();
        int sum = list.get(0);
        int i = 0;
        while (i < list.size()){
            if (i == list.size() - 1) {
                result.add(sum);
            } else if (list.get(i).equals(list.get(i + 1))) {
                sum += list.get(i);
            } else {
                result.add(sum);
                sum = list.get(i + 1);
            }
            i++;
        }
        return result;
    }
}

Test cases to validate our solution

import static org.junit.Assert.*;
import java.util.Arrays;
import java.util.List;
import org.junit.Test;

public class ConsecutivesTest {

    @Test
    public void test() {
        System.out.println("Basic Tests");
        List<Integer> i = Arrays.asList(1,4,4,4,0,4,3,3,1);
        List<Integer> o = Arrays.asList(1,12,0,4,6,1);
        System.out.println("Input: {1,4,4,4,0,4,3,3,1}");
        assertEquals(o, Consecutives.sumConsecutives(i));
        i = Arrays.asList(-5,-5,7,7,12,0);
        o = Arrays.asList(-10,14,12,0);
        System.out.println("Input: {-5,-5,7,7,12,0}");
        assertEquals(o, Consecutives.sumConsecutives(i));
        i = Arrays.asList(1,1,7,7,3);
        o = Arrays.asList(2,14,3);
        System.out.println("Input: {1,1,7,7,3}");
        assertEquals(o, Consecutives.sumConsecutives(i));
        i = Arrays.asList(3,3,3,3,1);
        o = Arrays.asList(12, 1);
        System.out.println("Input: {3,3,3,3,1}");
        assertEquals(o, Consecutives.sumConsecutives(i));
        i = Arrays.asList(2,2,-4,4,5,5,6,6,6,6,6,1);
        o = Arrays.asList(4, -4, 4, 10, 30, 1);
        System.out.println("Input: {2,2,-4,4,5,5,6,6,6,6,6,1}");
        assertEquals(o, Consecutives.sumConsecutives(i));
        i = Arrays.asList(1,1,1,1,1,3);
        o = Arrays.asList(5, 3);
        System.out.println("Input: {1,1,1,1,1,3}");
        assertEquals(o, Consecutives.sumConsecutives(i));
        i = Arrays.asList(1,-1,-2,2,3,-3,4,-4);
        o = Arrays.asList(1, -1, -2, 2, 3, -3, 4, -4);
        System.out.println("Input: {1,-1,-2,2,3,-3,4,-4}");
        assertEquals(o, Consecutives.sumConsecutives(i));
        
    }

}
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Andrew
Andrew

Andrew is a visionary software engineer and DevOps expert with a proven track record of delivering cutting-edge solutions that drive innovation at Ataiva.com. As a leader on numerous high-profile projects, Andrew brings his exceptional technical expertise and collaborative leadership skills to the table, fostering a culture of agility and excellence within the team. With a passion for architecting scalable systems, automating workflows, and empowering teams, Andrew is a sought-after authority in the field of software development and DevOps.

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