45_jump_game_ii.md

45 - Jump Game II

leetcode link

Given an array of non-negative integers, you are initially positioned at the first index of the array.

Each element in the array represents your maximum jump length at that position.

Your goal is to reach the last index in the minimum number of jumps.

Example:

Input: [2,3,1,1,4]
Output: 2
Explanation: The minimum number of jumps to reach the last index is 2.
    Jump 1 step from index 0 to 1, then 3 steps to the last index.

Note:

You can assume that you can always reach the last index.

Solution: Dynamic Programing + pruning - [O(n) - O(n^2)]

// 10/09/2020
// dp solution with pruning (maxreach)
int jump(vector<int>& nums) {
    vector<int> dp(nums.size(), nums.size());
    dp[0] = 0;
    int maxreach = 0;
    for(int i = 0, s = nums.size(); i != s; ++i){
        for(int j = maxreach+1; j <= i + nums[i] && j < s; ++j)
            dp[j] = min(dp[j], dp[i]+1);
        maxreach = i + nums[i];
    }
    return dp[nums.size()-1];
}

Gab's solution - O(n)

int jump(vector<int>& nums) {
    if (nums.size() <= 2) return nums.size() - 1;
    int dist = nums[0], max_dist = nums[0], jumps = 1;
    for (int i = 1; i != nums.size() - 1; ++i) {
        max_dist = max(--max_dist, nums[i]);
        if (--dist == 0) {
            dist = max_dist;
            ++jumps;
        }
    }
    return jumps;
}

Solutions that get TLE

naïve recursion - TLE

// 10/09/2020
// recursive solution 1 : TLE
// hard to do memoization with it
int jump(vector<int>& nums) {
    int minjumps = nums.size();
    jump(nums, minjumps);
    return minjumps;
}

void jump(vector<int> nums, int& minjumps, int i = 0, int nbjump = 0){
    if(i >= nums.size()-1) {
        minjumps = min(minjumps, nbjump);
        return;
    }
    for(int j = 1; j <= nums[i]; ++j)
        jump(nums, minjumps, i+j, nbjump+1);
}

simpler recursion - TLE

// 10/09/2020
// simpler recurive solution : TLE
int jump(vector<int>& nums, int i=0) {
    if(i >= nums.size()-1) return 0;
    int minjumps = nums.size();
    for(int j = 1; j <= nums[i]; ++j)
        minjumps = min(minjumps, 1 + jump(nums, i + j));
    return minjumps;
}

recursion + memoization - TLE :-(

// 10/09/2020
// recursion + memoization --> TLE
int jump(vector<int>& nums) {
    vector<int> memo(nums.size(), -1);
    return jump(nums, memo);
}

int jump(vector<int>& nums, vector<int>& memo, int i=0) {
    if(i == nums.size()-1) return 0;
    if(memo[i] == -1){
        int minjumps = nums.size();
        for(int j = 1, s = nums.size(); j <= nums[i] && i+j < s; ++j)
            minjumps = min(minjumps, 1 + jump(nums, memo, i + j));
        memo[i] = minjumps;
    }
    return memo[i];
}