Introduction Link to heading
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 Link to heading
Dynamic programming solution:
func jump(nums []int) int {
n := len(nums)
if n <= 1 {
return 0
}
dp := make([]int, n+1)
for i := 0; i <= n; i++ {
dp[i] = n+1
}
dp[0] = 0
j := nums[0]
for i, v := range nums {
if j < i {
// unachievable
return -1
}
j = max(j, i+v)
m := min(j, n-1)
for k := i; k <= m; k++ {
dp[k] = min(dp[k], dp[i] + 1)
}
}
return dp[n-1]
}
func min(a, b int) int {
if a < b {
return a
} else {
return b
}
}
func max(a, b int) int {
if a > b {
return a
} else {
return b
}
}
BFS solution
func jump(nums []int) int {
n := len(nums)
if n <= 1 {
return 0
}
steps := nums[0]
if steps >= n-1 {
return 1
}
maxSoFar, maxIndex := 0, 0
for i := 1; i <= steps; i++ {
nextSteps := i + nums[i]
if nextSteps >= n-1 {
// we achieved the last index
return 2
}
if maxSoFar < nextSteps {
maxSoFar = nextSteps
maxIndex = i
}
}
if maxIndex == 0 {
panic("no solution")
}
return 1 + jump(nums[maxIndex:])
}
func min(a, b int) int {
if a < b {
return a
} else {
return b
}
}
Optimized BFS solution
func jump(nums []int) int {
n := len(nums)
if n <= 1 {
return 0
}
steps := 1
i, m := 0, nums[0]
for m > 0 {
if m >= n-1 {
return steps
}
steps++
maxSoFar := 0
for ;i <= m; i++ {
maxSoFar = max(maxSoFar, i + nums[i])
if maxSoFar >= n-1 {
return steps
}
}
m = maxSoFar
}
return -1
}
func max(a, b int) int {
if a > b {
return a
} else {
return b
}
}
Explanation Link to heading
The first solution that come in mind is dynamic programming approach based on more simple version of this problem “Jump Game”. But, this solution is not optimal. There is another one based on BFS with recursion. It is also possible to optimize BFS solution for better performace by removing recursion. We already know the maxinum next index, all we need to continue the loop by incrementing steps.