By Long Luo
This article is the solution Leetcode Jump Game Problems Series.
55. Jump Game
Brute Force
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| public boolean canJump(int[] nums) {
int len = nums.length;
boolean[] visited = new boolean[len];
visited[0] = true;
for (int i = 0; i < len; i++) {
int steps = nums[i];
if (visited[i] && steps > 0) {
for (int j = 1; j <= steps && i + j < len; j++) {
visited[i + j] = true;
}
}
}
return visited[len - 1];
}
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Analysis
- Time Complexity: O(n2)
- Space Complexity: O(n)
Greedy
Jump to the farest postion.
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| public boolean canJump(int[] nums) {
int len = nums.length;
int maxIdx = 0;
for (int i = 0; i < len; i++) {
int steps = nums[i];
if (maxIdx >= i) {
maxIdx = Math.max(maxIdx, i + steps);
if (maxIdx >= len - 1) {
return true;
}
}
}
return false;
}
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Analysis
- Time Complexity: O(n)
- Space Complexity: O(1)
45. Jump Game II
DP
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| public static int jump_dp(int[] nums) {
int len = nums.length;
if (len <= 1) {
return 0;
}
int[] dp = new int[len];
Arrays.fill(dp, Integer.MAX_VALUE);
dp[0] = 0;
int minStep = Integer.MAX_VALUE;
for (int i = 0; i < len; i++) {
int num = nums[i];
for (int j = 1; j <= num && i + j < len; j++) {
dp[i + j] = Math.min(dp[i + j], dp[i] + 1);
if (i + j >= len - 1) {
minStep = Math.min(minStep, dp[i] + 1);
return minStep;
}
}
}
return minStep;
}
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Analysis
- Time Complexity: O(n2)
- Space Complexity: O(n)
Greedy
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| public int jump(int[] nums) {
int maxPosition = 0;
int end = 0;
int steps = 0;
for (int i = 0; i < nums.length - 1; i++) {
maxPosition = Math.max(maxPosition, i + nums[i]);
if (i == end) {
end = maxPosition;
steps++;
}
}
return steps;
}
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Think reverse, from destination to origin position.
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| public int jump(int[] nums) {
int ans = 0;
int position = nums.length - 1;
while (position > 0) {
for (int i = 0; i < position; i++) {
if (i + nums[i] >= position) {
ans++;
position = i;
break;
}
}
}
return ans;
}
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Analysis
- Time Complexity: O(n)
- Space Complexity: O(1)
1306. Jump Game III
BFS
BFS Solution:
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| public boolean canReach_bfs_opt(int[] arr, int start) {
if (arr[start] == 0) {
return true;
}
int len = arr.length;
boolean[] visited = new boolean[len];
Queue<Integer> queue = new LinkedList<>();
queue.offer(start);
visited[start] = true;
while (!queue.isEmpty()) {
int curPos = queue.poll();
if (arr[curPos] == 0) {
return true;
}
int right = curPos + arr[curPos];
if (right >= 0 && right < len && !visited[right]) {
visited[right] = true;
queue.offer(right);
}
int left = curPos - arr[curPos];
if (left >= 0 && left < len && !visited[left]) {
visited[left] = true;
queue.offer(left);
}
}
return false;
}
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Analysis
- Time Complexity: O(n)
- Space Complexity: O(n)
DFS
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| public boolean canReach_dfs(int[] arr, int start) {
int len = arr.length;
boolean[] vis = new boolean[len];
return dfs(arr, vis, start);
}
public boolean dfs(int[] arr, boolean[] vis, int start) {
if (start < 0 || start >= arr.length || vis[start]) {
return false;
}
vis[start] = true;
if (arr[start] == 0) {
return true;
}
return dfs(arr, vis, start - arr[start]) || dfs(arr, vis, start + arr[start]);
}
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Analysis
- Time Complexity: O(n)
- Space Complexity: O(n)
1345. Jump Game IV
BFS
Analysis
- Time Complexity: O(n)
- Space Complexity: O(n)
1340. Jump Game V
Analysis
- Time Complexity: O(n)
- Space Complexity: O(n)
1696. Jump Game VI
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| public static boolean canJump(int[] nums) {
int len = nums.length;
int maxIdx = 0;
for (int i = 0; i < len; i++) {
int steps = nums[i];
if (maxIdx >= i) {
maxIdx = Math.max(maxIdx, i + steps);
if (maxIdx >= len - 1) {
return true;
}
}
}
return false;
}
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Analysis
- Time Complexity: O(n)
- Space Complexity: O(n)
1871. Jump Game VII
Analysis
- Time Complexity: O(n)
- Space Complexity: O(n)
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