Step 1:- [Defining the States of DP]

Here state of dp[ i ] is maximum amount that can be robbed from 0 to i-th houses and at the same time i-th house must be robbed.

Step 2:- [Identifing the Base Case for the solution]

Initializing dp[ 0 ], dp[ 1 ], dp[ 2 ] values.

Step 3 :- [Identifing the Relation between the dp States and framing an Equation]

//robbing i-th house can be done by 2 case,

case 1 : just neglecting i-1 th house and rob i-th house (neglecting last one house) OR case 2 : neglecting both i-1 th house, i-2 th house and rob i-th house (neglecting last 2 houses)

dp[ i ] = max( dp[ i - 2 ], dp[ i - 3 ] ) + nums[ i ] ;

BECAUSE for example consider 100,1,2,100. then answer = (100 + 100) 0-th and 3-rd houses are robbed. for robbing 3rd house (i.e for dp[ 3 ]) ,here, neglecting last 2 houses leads to answer[200].

class Solution {
public:
    int rob(vector<int>& nums) {
        int n=nums.size();

        if(n==0) return 0;  //if their are no houses

            else if(n==1) return nums[0];	//only one house

                else if(n==2) return max(nums[0],nums[1]);   //only two houses

        int dp[n];  //for more than two houses

        dp[0]=nums[0];  dp[1]=nums[1];  dp[2]=nums[0]+nums[2];

        for(int i=3;i<n;i++)
        {
            if(dp[i-2]>dp[i-3])     //neglecting only 1 previous house.
                dp[i] = dp[i-2] + nums[i];
            else                          //neglecting previous 2 houses (as explained in example).
                dp[i] = dp[i-3] + nums[i];
        }
            return max(dp[n-1],dp[n-2]);
    }
};