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Program to find maximum sum of two sets where sums are equal 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
Updated on: 12-Dec-2020
74 Views
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