A naive solution

Lecture #2: Two Sample Problems in Analysis

A naive solution

Compute (and store) sum for each subsequence we can use an n x n array sum to store the values for the sums of all possible subsequences from A[i] to A[j], putting the value in the cell sum[i,j]

Q: what part of the array will we fill in?

Lecture #2: Two Sample Problems in Analysis

A naive solution

	Compute (and store) sum for each subsequence
	Finding the result to return the correct answer, we need only fill in the array,then search it for the maximum value Q: what is the asymptotic running time for this approach (in O notation)?

Lecture #2: Two Sample Problems in Analysis

A naive solution

	Compute (and store) sum for each subsequence
	Finding the result
	Keeping a running maximum we can eliminate the search by keeping and updating a "current maximum value" as we fill in the array Q: does eliminating this (inefficient) search help our asymptotic running time?

Lecture #2: Two Sample Problems in Analysis

A naive solution

	Compute (and store) sum for each subsequence
	Finding the result
	Keeping a running maximum
	Cutting down on space once we use a running maximum, we need not actually build the array Q: does eliminating the array of maxima help our asymptotic space usage? Q: what other "running values" do we need to keep track of?

Lecture #2: Two Sample Problems in Analysis

A naive solution

	Compute (and store) sum for each subsequence
	Finding the result
	Keeping a running maximum
	Cutting down on space
	Code for the naive algorithm (cubic) /** * Cubic maximum contiguous subsequence sum algorithm. * seqStart and seqEnd represent the actual best sequence. */ public static int maxSubSum1( int [ ] a ) { int maxSum = 0; for( int i = 0; i < a.length; i++ ) for( int j = i; j < a.length; j++ ) { int thisSum = 0; for( int k = i; k <= j; k++ ) thisSum += a[ k ]; if( thisSum > maxSum ) { maxSum = thisSum; seqStart = i; seqEnd = j; } } return maxSum; }