Program to find minimum distance to the target element using Python

We need to find the minimum distance from a start position to any occurrence of a target element in an array. The distance is calculated as the absolute difference between indices.

Given an array nums, a target value that exists in the array, and a start index, we find the index i where nums[i] = target and |i - start| is minimized.

Example

If nums = [3,4,5,6,7], target = 7, and start = 2, the target 7 is found at index 4. The distance is |4 - 2| = 2.

Algorithm

The approach is straightforward ?

• Initialize minimum to infinity

• Iterate through the array with index i

• If nums[i] equals target, calculate |i - start|

• Update minimum if current distance is smaller

• Return the minimum distance found

Implementation

from math import inf

def solve(nums, target, start):
    minimum = inf
    for i in range(len(nums)):
        if nums[i] == target:
            if abs(i - start) < minimum:
                minimum = abs(i - start)
    return minimum

nums = [3, 4, 5, 6, 7]
target = 7
start = 2
print(solve(nums, target, start))

2

Optimized Approach

We can optimize by using Python's built-in min() function with a generator expression ?

def solve_optimized(nums, target, start):
    return min(abs(i - start) for i, val in enumerate(nums) if val == target)

nums = [3, 4, 5, 6, 7]
target = 7
start = 2
print(solve_optimized(nums, target, start))

2

Multiple Target Occurrences

Let's test with an array containing multiple occurrences of the target ?

def solve(nums, target, start):
    minimum = float('inf')
    for i in range(len(nums)):
        if nums[i] == target:
            distance = abs(i - start)
            if distance < minimum:
                minimum = distance
    return minimum

nums = [1, 0, 3, 5, 9, 5]
target = 5
start = 2
print(f"Array: {nums}")
print(f"Target: {target}, Start: {start}")
print(f"Minimum distance: {solve(nums, target, start)}")

Array: [1, 0, 3, 5, 9, 5]
Target: 5, Start: 2
Minimum distance: 1

The target 5 appears at indices 3 and 5. From start position 2, the distances are |3-2| = 1 and |5-2| = 3. The minimum is 1.

Conclusion

This algorithm finds the minimum distance to a target element by iterating through the array once, giving O(n) time complexity. The optimized version using min() is more concise but has the same time complexity.