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55. jump game

admin 11月 13, 2020 0

Question

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.

Determine if you are able to reach the last index.

Analysis

Start with the recursive backtracking solution
Optimize by using a memoization table (top-down[3] dynamic programming)
Remove the need for recursion (bottom-up dynamic programming)
Apply final tricks to reduce the time / memory complexity

Complexity

Time: O(2^n) Space: O(n) recursion requires additional memory.
Time: O(n^2) Space: O(2n)
Time: O(n^2) Space: O(n)
Time: O(n) Space: O(1)

Solution

enum Index {
    GOOD, BAD, UNKNOWN
}
class  {
    
    

    public boolean canJump(int[] nums) {
        return canJumpFromPosition(0, nums);
    }
    
    
    public boolean canJumpFromPosition(int position, int[] nums) {
        if (position == nums.length - 1) {
            return true;
        }
        
        int furthestJump = Math.min(position + nums[position], nums.length - 1);
        for (int nextPosition = furthestJump; nextPosition > position; nextPosition--) {
            if (canJumpFromPosition(nextPosition, nums)) {
                return true;
            }
        }
        return false;
    }
    */
    
    // TLE Appraoch #2 Dynamic Programming Top-down Memoization 
    /*
    Index[] memo;
    
    public boolean canJumpFromPosition(int position, int[] nums) {
        if (memo[position] != Index.UNKNOWN) {
            return memo[position] == Index.GOOD ? true : false;
        }
        
        int furthestJump = Math.min(position + nums[position], nums.length - 1);
        for (int nextPosition = position + 1; nextPosition <= furthestJump; nextPosition++) {
            if (canJumpFromPosition(nextPosition, nums)) {
                memo[position] = Index.GOOD;
                return true;
            }
        }
        
        memo[position] = Index.BAD;
        return false;
    }
    
    public boolean canJump(int[] nums) {
        memo = new Index[nums.length];
        for (int i = 0; i < memo.length; i++) {
            memo[i] = Index.UNKNOWN;
        }
        
        memo[memo.length - 1] = Index.GOOD;
        return canJumpFromPosition(0, nums);
    }
    */
    
    
    // AC Approach #3 (Dynamic Programming Bottom-up)
    /*
    public boolean canJump(int[] nums) {
        Index[] memo = new Index[nums.length];
        for (int i = 0; i < memo.length; i++) {
            memo[i] = Index.UNKNOWN;
        }
        memo[memo.length - 1] = Index.GOOD;
        
        for (int i = nums.length - 2; i >= 0; i--) {
            int furthestJump = Math.min(i + nums[i], nums.length - 1);
            for (int j = i + 1; j <= furthestJump; j++) {
                if (memo[j] == Index.GOOD) {
                    memo[i] = Index.GOOD;
                    break;
                }
            }
        }
        return memo[0] == Index.GOOD;
    }
    */
    
    public boolean canJump(int[] nums) {
        int lastPos = nums.length - 1;
        for (int i = nums.length - 1; i >= 0; i--) {
            if (i + nums[i] >= lastPos) {
                lastPos = i;
            }
        }
        
        return lastPos == 0;
    }
}