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Topic: Longest SubSequence with Sum <= K (Read 6222 times) |
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sateesp
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Longest SubSequence with Sum <= K
« on: Sep 4th, 2010, 12:20pm » |
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Given an array, find the longest sub set whose sum is less or equal then the given MaxSum int[] FindMaxSumArray(int[] array, int maxsum) for example, given array: {1, -2, 4, 5, -2, 6, 7} maxsum=7 The result would be: {1,-2, -2, 6}
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towr
wu::riddles Moderator Uberpuzzler
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Re: Longest SubSequence with Sum <= K
« Reply #1 on: Sep 4th, 2010, 1:08pm » |
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Sort the array, and the take the smallest numbers that sum to less than or equal the maxsum.
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Grimbal
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Re: Longest SubSequence with Sum <= K
« Reply #2 on: Sep 8th, 2010, 5:13am » |
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The *longest* subset? Wouldn't {1, -2, 4, 5, -2} (sum 6) be longer?
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birbal
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Re: Longest SubSequence with Sum <= K
« Reply #3 on: Sep 11th, 2010, 2:04am » |
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on Sep 8th, 2010, 5:13am, Grimbal wrote:The *longest* subset? Wouldn't {1, -2, 4, 5, -2} (sum 6) be longer? |
| Yes, i think {1, -2, 4, 5, -2} this should be the set.
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MrLambert
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Re: Longest SubSequence with Sum <= K
« Reply #4 on: Oct 31st, 2011, 9:40am » |
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Yes, its a mistake set should be: {1, -2, 4, 5, -2}
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computing
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Re: Longest SubSequence with Sum <= K
« Reply #5 on: Nov 7th, 2011, 2:01am » |
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One simple way is to use Kadane's algorithm and keep verifying input constraints, in such a way that the sum_so_far should be less than MaxSum, if so, consider next element. Otherwise, current indices [i, j] of sum_so_far will be answer. A DP solution. We may need to repeat the procedure for all those sub array sizes that are less than current sub array that satisfies input constraints. I mean, we also need to check the remaining array [j+1, n], if (n - j) > (j - i + 1). Kadane algorithm - http://en.wikipedia.org/wiki/Maximum_subarray_problem Any thoughts?
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« Last Edit: Nov 7th, 2011, 2:02am by computing » |
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kaka
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Re: Longest SubSequence with Sum <= K
« Reply #6 on: Nov 28th, 2011, 11:40pm » |
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@computing: Kadane's algorithm is about finding the contiguous subarray where as this one is about sub set (order doesn't matter)
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