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Maximum Sum Increasing Subsequence
Maximum Sum Increasing subsequence is a subsequence of a given list of integers, whose sum is maximum and in the subsequence, all elements are sorted in increasing order.
Let there is an array to store max sum increasing subsequence, such that L[i] is the max sum increasing subsequence, which is ending with array[i].
Input and Output
Input:
Sequence of integers. {3, 2, 6, 4, 5, 1}
Output:
Increasing subsequence whose sum is maximum. {3, 4, 5}.Algorithm
maxSumSubSeq(array, n)
Input: The sequence of numbers, number of elements.
Output: Maximum sum of the increasing sub sequence.
Begin define array of arrays named subSeqLen of size n. add arr[0] into the subSeqLen for i in range (1 to n-1), do for j in range (0 to i-1), do if arr[i] > arr[j] and sum of subSeqLen [i] < sum of subSeqLen [j], then subSeqLen[i] := subSeqLen[j] done done add arr[i] into subSeqLen[i] res := subSeqLen[0] for all values of subSeqLen, do if sum of subSeqLen[i] > sum of subSeqLen[res], then res := subSeqLen[i] done print the values of res. End
Example
#include <iostream>
#include <vector>
using namespace std;
int findAllSum(vector<int> arr) { //find sum of all vector elements
int sum = 0;
for(int i = 0; i<arr.size(); i++) {
sum += arr[i];
}
return sum;
}
void maxSumSubSeq(int arr[], int n) {
vector <vector<int> > subSeqLen(n); //max sum increasing subsequence ending with arr[i]
subSeqLen[0].push_back(arr[0]);
for (int i = 1; i < n; i++) { //from index 1 to all
for (int j = 0; j < i; j++) { //for all j, j<i
if ((arr[i] > arr[j]) && (findAllSum(subSeqLen[i]) < findAllSum(subSeqLen[j])))
subSeqLen[i] = subSeqLen[j];
}
subSeqLen[i].push_back(arr[i]); //sub Sequence ends with arr[i]
}
vector<int> res = subSeqLen[0];
for(int i = 0; i<subSeqLen.size(); i++) {
if (findAllSum(subSeqLen[i]) > findAllSum(res))
res = subSeqLen[i];
}
for(int i = 0; i<res.size(); i++)
cout << res[i] << " ";
cout << endl;
}
int main() {
int arr[] = { 3, 2, 6, 4, 5, 1 };
int n = 6;
cout << "The Maximum Sum Subsequence is: ";
maxSumSubSeq(arr, n);
}Output
The Maximum Sum Subsequence is: 3 4 5
Updated on: 17-Jun-2020
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