Delete and Earn in C++

Suppose we have an array nums of integers, we can perform some operations on the array. Here in each operation, we pick any nums[i] and delete it to earn nums[i] amount of points. We must delete every element equal to nums[i] - 1 or nums[i] + 1. Initially the point is 0. We have to find the maximum number of points we can earn by applying such operations. So if the input is like [3,4,2], then the output will be 6. So this is because, if we delete 4, we will get 4 points, consequently 3 will also be deleted. Then delete 2 to get 2 points. 6 total points are earned.

To solve this, we will follow these steps −

• n := size of nums array, define map m, ret := 0, store the frequency of elements in nums into m

• cnt := 0

• for each pair it of m

  • x := key of it

  • temp := x * value of it

  • it1 := point to previous of it, and it2 := point to the previous of it1

  • if cnt >= 1 and x – key of it1 > 1, then temp := m[key of it1]

  • otherwise when cnt >= 2, then temp := temp + m[key of it2]

  • a = m[key of it1] if cnt >= 1, otherwise 0

  • m[key of it] := max of temp and a

  • ret := max of ret and temp

  • increase cnt by 1

• return ret

Example(C++)

Let us see the following implementation to get a better understanding −

 Live Demo

#include <bits/stdc++.h> using namespace std; class Solution { public:    int deleteAndEarn(vector<int>& nums) {       int n = nums.size();       map <int, int> m;       int ret = 0;       for(int i = 0; i < nums.size(); i++){          m[nums[i]]++;       }       int cnt = 0;       map <int, int> :: iterator it = m.begin();       while(it != m.end()){          int x = it->first;          int temp = x * it->second;          map <int, int> :: iterator it1 = prev(it);          map <int, int> :: iterator it2 = prev(it1);          if(cnt >= 1 && x - it1->first > 1){             temp += m[it1->first];          }          else if(cnt >= 2){             temp += m[it2->first];          }          m[it->first] = max(temp, cnt >= 1 ? m[it1->first] : 0);          ret = max(ret, temp);          it++;          cnt++;       }       return ret;    } }; main(){    vector<int> v = {3,4,2};    Solution ob;    cout << (ob.deleteAndEarn(v)); }

Input

[3,4,2]

Output

6