Striver's SDE Sheet Journey - 4 Kadane's Algorithm

HiDevs,

In the previous post, we have solved and understood the Next Permutation problem and in this post, we will tackle the next one Kadanes algorithms.

#4 Kadane's Algorithms

in this problem we have given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.

Example 1:

Input: nums = [-2,1,-3,4,-1,2,1,-5,4]
Output: 6
Explanation: [4,-1,2,1] has the largest sum = 6.

Example 2:

Input: nums = [1]
Output: 1

Solution

we can easily solve this problem using the kadane's algorithms.
lets discuss Kadane's algorithms step by step.

step-1 initialize three int variable sum = 0 ,max = nums[0], arrSize = nums.length.

step-2 run a loop from i = 0 to arrSize.

1. sum = sum + nums[i].
2. if sum > max then max = sum.
3. if sum < 0 then sum = 0.

step-3 return max

step-4 end

why reinitialize sum = 0 if sum < 0?
negative-sum never contribute to maximum sum, so it is better to reinitialize sum to 0 and start adding from next element

before coding this algorithm in java, have a look at this image for a better understanding of this algo.

Java

class Solution {    public int maxSubArray(int[] nums) {        int sum = 0;        int max = nums[0];        int arrSize = nums.length;        for(int i=0; i < arrSize; i++){            sum = sum + nums[i];            if(sum > max){                max = sum;            }            if(sum < 0) {                sum =0            }        }        return max;}}

But how? can we know,

• length of the max subarray

• start & end index of the max subarray

• elements of the max subarray

so, for these, we can make two more int variables firstIndex & lastIndex that store the start & last index number of the max subarray and then do modification in the step-2.

step-2 run a loop from i = 0 to arrSize.

1. sum = sum + nums[i].
2. if sum > max then max = sum,lastIndex = i.
3. if sum < 0 then sum = 0, firstIndex = i+1.

using firstIndex and lastIndex variables, now we can answer the following questions.

• length of the subarray.
  subArrSize = lastIndex - firstIndex

• start & end index of subarray.
  by using firstIndex and lastIndex variable

• elements of the max subarray.
  by traversing from firstIndex to lastIndex we can find the elements

Java

class Solution {    public int maxSubArray(int[] nums) {        int sum = 0;        int max = nums[0];        int arrSize = nums.length;        int firstIndex = 0,lastIndex = 0;        for(int i=0; i < arrSize; i++){            sum = sum + nums[i];            if(sum > max){                max = sum;                lastIndex = i;            }            if(sum < 0) {                sum =0;                firstIndex = i + 1;            }        }        return max;}}

Time Complexity

running a for loop from 0 to arrSize.
so, O(arrSize) will be it's time complexity.

Space Complexity

the algo is not using any extra space, then O(1) will be its space complexity.

if you like this article please let me in the comment section.
if you find anything wrong in the post,plz correct me.
Thank you for reading.