Dynamic-Programming/55. Jump Game at main · VaspulaVijayaLakshmi/Dynamic-Programming · GitHub

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// TLE
class Solution {
public:

    bool canJump(vector<int>& nums, int idx){

        if(idx>=nums.size()-1)
         return true;

       int maxSteps = nums[idx];
       for(int i=1;i<=maxSteps;i++){

         if(canJump(nums,idx+i))
          return true;
       }

       return false;

    }


    bool canJump(vector<int>& nums) {
       return canJump(nums,0);
    }
};


MEMO

// TLE
class Solution {
public:

    bool canJump(vector<int>& nums, int idx,vector<bool> &dp){

        if(idx>=nums.size()-1)
         return true;

       if(dp[idx]==true)
       return dp[idx];

       int maxSteps = nums[idx];
       for(int i=1;i<=maxSteps;i++){

         if(canJump(nums,idx+i,dp))
          return dp[idx] =  true;
       }

       return dp[idx]=false;

    }


    bool canJump(vector<int>& nums) {
        vector<bool> dp(nums.size());
       return canJump(nums,0,dp);
    }
};


GREEDY

class Solution {
public:

    bool canJump(vector<int>& nums) {

        int minJumps=0;
        for(int i=nums.size()-2;i>=0;i--){

            minJumps++;

            //here we should not return, because
            //we may come across some big number in the front, so tht we may not need to jump on this step
            // if(nums[i]<minJumps)
            //  return false;


// also if we know with the prev number we r able to land on last idx, then we just need to check how we can reach the curr idx, not to end

            if(nums[i]>=minJumps)
              minJumps=0;

        }

        return minJumps==0;

    }
};