Very elegant C++ solution, O(N) time, O(N) space, using DP and pruning

  • 0
     class Solution {
            	int jump(vector<int>& nums) {
            		if(nums.empty())    return -1;
            		int N = nums.size(),lastFarthest=0;  	        
            		vector<int> dp(N,N+1);
            		for(int i=0;i<N-1;i++){
            			int cover_length = nums[i];//cover range:i ~ i+cover_length;
            			if(i+cover_length <=lastFarthest)   continue;  //pruning
            			if(i+cover_length >=N-1)	
            				return 1+dp[i];          //pruning
            			for(int j=lastFarthest+1;j<=i+cover_length && j<N;j++){
            				dp[j] = 1+dp[i];
            			lastFarthest = i+cover_length;
            		return dp[N-1];
        //nums:[2,3,1,1,4,1,2,3,1,2,1 ,2 ,1 ,3]
        //dp:  [0,1,1,2,2,3,3,3,3,4,4 ,5, 6 ,6]

  • 0

    I think the time complexity of your algorithm is O(n^2) rather than O(n), but the pruning strategy is very good!

  • 0

    well, I think it is O(N) because of prunning, each element in dp vector only need one time's calculation.

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