PROBLEM DESCRIPTION
You are given a 0-indexed array of integers nums of length n. You are initially positioned at nums[0].
Each element nums[i] represents the maximum length of a forward jump from index i. In other words, if you are at nums[i], you can jump to any nums[i + j] where:
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0 <= j <= nums[i] and
i + j < n
Return the minimum number of jumps to reach nums[n - 1]. The test cases are generated such that you can reach nums[n - 1].
SOLUTION
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class Solution {
public int jump(int[] nums) {
int n = nums.length;
//current goal
int current = n-1;
//total jumps
int jumps = 0;
//while we have not reached the start position which is index 0
while(current != 0){
//since we are not at 0, at least one more jump is needed
jumps++;
//init for next goal
int next = current-1;
//get the left most index from which we can reach the current goal
for(int i=current-1; i>=0; i--){
//if current position + max jump possible is >= current goal, it's a possible next goal
//continue to check for other left position
//finally, we will have the left most index from which we can reach the current goal
if(i + nums[i] >= current){
next = i;
}
}
//update the next goal
current = next;
}
return jumps;
}
}
SECOND APPROACH (SIMPLIFIED BFS)
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class Solution {
public int jump(int[] nums) {
int n = nums.length;
int jumps = 0;
//window start to end to store the start and end of a "level/window".
int l=0, r=0;
//until we are at end index, keep updating L and R (next level indices)
while(r<n-1){
//since we've not reached the destination yet, at least one more jump is needed
jumps++;
//init
int maxJumpPossible = r+1;
//check the max jump possible for the current windows
for(int i=l; i<=r; i++){
maxJumpPossible = Math.max(maxJumpPossible, i+nums[i]);
}
//next window/level will start from (R+1) to (maxJumpPossible from the previous level)
l=r+1;
r = maxJumpPossible;
}
return jumps;
}
}