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Find Maximum Sum of Two Equal Sets in C++
Suppose we have a list of numbers called nums, now find two sets as their sums are same and maximum, then find sum value.
So, if the input is like nums = [2, 5, 4, 6], then the output will be 6, as the sets are [2, 4] and [6].
To solve this, we will follow these steps −
- sum := 0
- for each number i in nums, do
- sum := sum + i
- n := size of nums
- Define one 2D array dp of size (n + 1) x (2 * sum + 5) and fill with -1
- dp[0, sum] := 0
- for initialize i := 1, when i <= n, update (increase i by 1), do −
- x := nums[i - 1]
- for initialize j := 0, when j < 2 * sum + 5, update (increase j by 1), do −
- if j - x >= 0 and dp[i - 1, j - x] is not equal to -1, then ^−
- dp[i, j] := maximum of dp[i, j] and (dp[i - 1, j - x] + x)
- if j + x < (2 * sum + 5) and dp[i - 1, j + x] is not equal to -1, then −
- dp[i, j] := maximum of dp[i, j] and (dp[i - 1, j + x])
- dp[i, j] := maximum of dp[i, j] and dp[i - 1, j]
- if j - x >= 0 and dp[i - 1, j - x] is not equal to -1, then ^−
- return dp[n, sum]
Example (C++)
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h> using namespace std; class Solution { public: int solve(vector<int>& nums) { int sum = 0; for (int i : nums) sum += i; int n = nums.size(); vector<vector<int> > dp(n + 1, vector<int>(2 * sum + 5, -1)); dp[0][sum] = 0; for (int i = 1; i <= n; i++) { int x = nums[i - 1]; for (int j = 0; j < 2 * sum + 5; j++) { if (j - x >= 0 && dp[i - 1][j - x] != -1) { dp[i][j] = max(dp[i][j], dp[i - 1][j - x] + x); } if (j + x < 2 * sum + 5 && dp[i - 1][j + x] != -1) { dp[i][j] = max(dp[i][j], dp[i - 1][j + x]); } dp[i][j] = max(dp[i][j], dp[i - 1][j]); } } return dp[n][sum]; } }; int solve(vector<int>& nums) { return (new Solution())->solve(nums); } main(){ vector<int> v = {2, 5, 4, 6}; cout << solve(v); }
Input
{2, 5, 4, 6}
Output
6
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