Find the minimum of maximum length of a jump required to reach the last island in exactly k jumps in Python

Suppose we have an array A of numbers, where the i-th number represents the position of an island. Given another integer k (1 ? k

For example, if A = [7, 20, 41, 48] and k = 2, the output will be 28. There are two possible paths:

• Path 1: 7 ? 20 ? 48 (jumps: 13, 28) ? maximum jump = 28

• Path 2: 7 ? 41 ? 48 (jumps: 34, 7) ? maximum jump = 34

The minimum of these maximum jumps is 28.

Algorithm Overview

We use binary search on the answer. For each potential maximum jump distance, we check if it's possible to reach the destination in exactly k jumps using a greedy approach.

Helper Function: isPossible

This function checks if we can reach the last island with a given maximum jump distance ?

def isPossible(arr, dist, k):
    n = len(arr)
    req = 0  # Required jumps
    current = 0
    previous = 0
    
    for i in range(n):
        # Find the farthest island we can reach with current jump distance
        while current != n and (arr[current] - arr[previous]) <= dist:
            current += 1
        
        req += 1
        if current == n:
            break
        
        previous = current - 1
    
    # Check if we couldn't reach the end or need too many jumps
    if current != n:
        return False
    if req <= k:
        return True
    return False

Main Solution: Binary Search

We binary search on the maximum jump distance to find the minimum possible value ?

def minimum_distance(arr, k):
    n = len(arr)
    left = 0
    right = arr[-1]  # Maximum possible jump distance
    ans = 0
    
    while left <= right:
        mid = (left + right) // 2
        
        if isPossible(arr, mid, k):
            ans = mid
            right = mid - 1  # Try smaller distance
        else:
            left = mid + 1   # Need larger distance
    
    return ans

# Test the solution
arr = [7, 20, 41, 48]
k = 2
result = minimum_distance(arr, k)
print(f"Minimum maximum jump distance: {result}")

Minimum maximum jump distance: 28

How It Works

The algorithm works in two phases:

1. Binary Search: We search for the minimum possible maximum jump distance between 0 and the total distance.

2. Feasibility Check: For each candidate distance, we use a greedy approach to check if we can complete the journey in at most k jumps.

Complete Example

def isPossible(arr, dist, k):
    n = len(arr)
    req = 0
    current = 0
    previous = 0
    
    for i in range(n):
        while current != n and (arr[current] - arr[previous]) <= dist:
            current += 1
        req += 1
        if current == n:
            break
        previous = current - 1
    
    if current != n:
        return False
    if req <= k:
        return True
    return False

def minimum_distance(arr, k):
    n = len(arr)
    left = 0
    right = arr[-1]
    ans = 0
    
    while left <= right:
        mid = (left + right) // 2
        if isPossible(arr, mid, k):
            ans = mid
            right = mid - 1
        else:
            left = mid + 1
    return ans

# Test with different examples
test_cases = [
    ([7, 20, 41, 48], 2),
    ([1, 3, 6, 10], 3),
    ([0, 5, 15, 25], 2)
]

for islands, jumps in test_cases:
    result = minimum_distance(islands, jumps)
    print(f"Islands: {islands}, Jumps: {jumps} ? Result: {result}")

Islands: [7, 20, 41, 48], Jumps: 2 ? Result: 28
Islands: [1, 3, 6, 10], Jumps: 3 ? Result: 4
Islands: [0, 5, 15, 25], Jumps: 2 ? Result: 20

Time Complexity

The time complexity is O(n log(max_distance)), where n is the number of islands. The binary search runs in O(log(max_distance)) iterations, and each feasibility check takes O(n) time.

Conclusion

This problem combines binary search with greedy algorithms to find the optimal solution. The key insight is to binary search on the answer while using a greedy approach to check feasibility for each candidate maximum jump distance.