Kadane's continuous subarray algorithm

Kadane's continuous subarray algorithm

• 1.
  Kadane’s continuous subarrayalgorithm rerun.me by Arun Manivannan
• 2.
  Maximum sum subarrayproblem - find the contiguous subarray who has the maximum sum all you need to do is need to { find me -1 3 -5 4 6 -1 2 -7 13 -3 0 1 2 3 4 5 6 7 8 9 rerun.me
• 3.
  This program introducesfew variables on top of the default implementation of Kadane’s algorithm. We’ll have four major variables in all 1) cumulative sum - holds the cumulative sum of the array 2) maximum sum - holds the maximum sum of continuous items in the array. 3) maximum start index - start index of the sub array whose total is the maximum within the array 4) maximum end index - end index of the sub array whose total is the maximum within the array rerun.me
• 4.
  Let’s take asample bag of numbers stored in an array -1 3 -5 4 6 -1 2 -7 13 -3 0 1 2 3 4 5 6 7 8 9 index rerun.me
• 5.
  cum sum 0 max sum Integer.MIN_VALUE -1 3 -5 4 6 -1 2 -7 13 -3 0 1 2 3 4 5 6 7 8 9 current index 0 lower index 0 end index 0 rerun.me
• 6.
  cum sum -1 Keep incrementing our cumulative sum max sum -1 -1 3 -5 4 6 -1 2 -7 13 -3 0 1 2 3 4 5 6 7 8 9 current index 0 startIndex 0 end index 0 maxStartIndexUntilNow -1 rerun.me
• 7.
  When the cumulativesum is greater than max sum, it just means that the next number is an ‘useful addition’. So, lets set our max sum as our current cumulative sum, startIndex as the maxStartIndexUntilNow and cum sum 2 endIndex as our counter. prev max sum -1 max sum 2 if(cumulativeSum>maxSum){ maxSum = cumulativeSum; maxStartIndex=maxStartIndexUntilNow; maxEndIndex = currentIndex; } -1 3 -5 4 6 -1 2 -7 13 -3 0 1 2 3 4 5 6 7 8 9 current index 1 startIndex 0 end index 1 maxStartIndexUntilNow 0 rerun.me
• 8.
  When the cumulativesum is <0, it means that the next number is negative and actually bringing down the total sum. So, reset the cumulative sum as zero to restart the accumulation process from next number. Also, let’s set the fresh maxStartIndexUntilNow to the next number to reflect fresh accumulation actual cum sum -3 cum sum 0 else if (cumulativeSum<0){ max sum 2 maxStartIndexUntilNow=currentIndex+1; cumulativeSum=0; } -1 3 -5 4 6 -1 2 -7 13 -3 0 1 2 3 4 5 6 7 8 9 current index 2 That said, don’t bother reseting the endIndex or startIndex 0 startIndex or the maxSum. Instead save them aside since end index 1 there could be no series in the future which has a sum more than the one thus far. maxStartIndexUntilNow 3 rerun.me
• 9.
  Ahaa, looks likeour new cumulative sum is more than our max sum - let’s set our new max sum, new startIndex and end Index. our new ‘max’ end index would simply be our current Index. Our startIndex is just our ‘saved away’ startIndex cum sum 4 if(cumulativeSum>maxSum){ prev max sum 2 max sum 4 maxSum = cumulativeSum; maxStartIndex=maxStartIndexUntilNow; maxEndIndex = currentIndex; } -1 3 -5 4 6 -1 2 -7 13 -3 0 1 2 3 4 5 6 7 8 9 current index 3 prev startIndex 0 startIndex 3 prev endIndex 1 end index 3 maxStartIndexUntilNow 3 rerun.me
• 10.
  More Good newscoming along. cumulative sum is increasing and is getting more than the max sum every iteration. As before, reset max sum, end Index and startIndex cum sum 10 max sum 10 -1 3 -5 4 6 -1 2 -7 13 -3 0 1 2 3 4 5 6 7 8 9 current index 4 startIndex 3 end index 4 maxStartIndexUntilNow 3 rerun.me
• 11.
  A little losshere. However, our max sum is stronger than the cumulative sum still. So, let’s just ignore it and do nothing. cum sum 9 max sum 10 -1 3 -5 4 6 -1 2 -7 13 -3 0 1 2 3 4 5 6 7 8 9 current index 5 startIndex 3 end index 4 maxStartIndexUntilNow 3 rerun.me
• 12.
  Let’s gobble upthe next increment to the sum. Make max sum and end index reflect it. prev cum sum 9 cum sum 11 max sum 11 -1 3 -5 4 6 -1 2 -7 13 -3 0 1 2 3 4 5 6 7 8 9 current index 6 startIndex 3 end index 6 maxStartIndexUntilNow 3 rerun.me
• 13.
  A setback herebut we have our savings - max indices and sum. Let’s pretend that this never happened. prev cum sum 11 cum sum 4 max sum 11 -1 3 -5 4 6 -1 2 -7 13 -3 0 1 2 3 4 5 6 7 8 9 current index 7 startIndex 3 end index 6 maxStartIndexUntilNow 3 rerun.me
• 14.
  Goodness !!! cum sum 17 max sum 17 -1 3 -5 4 6 -1 2 -7 13 -3 0 1 2 3 4 5 6 7 8 9 current index 8 startIndex 3 end index 8 maxStartIndexUntilNow 3 rerun.me
• 15.
  A final drop.But hey, we’ve passed the maximum in the last pass. Let’s ignore this pass and return the previous results. cum sum 14 max sum 17 -1 3 -5 4 6 -1 2 -7 13 -3 0 1 2 3 4 5 6 7 8 9 current index 9 The contiguous array which has the maximum sum is startIndex 3 sandwiched within the 3rd and the 8th indices. end index 8 The maximum sum of the contiguous array is 17 maxStartIndexUntilNow 3 rerun.me