Fix: Ensure all skills are tracked as files, not submodules

skills/loki-mode/benchmarks/results/2026-01-05-00-49-17/humaneval-solutions/129.py

@@ -0,0 +1,61 @@
+def minPath(grid, k):
+    """
+    Given a grid with N rows and N columns (N >= 2) and a positive integer k, 
+    each cell of the grid contains a value. Every integer in the range [1, N * N]
+    inclusive appears exactly once on the cells of the grid.
+
+    You have to find the minimum path of length k in the grid. You can start
+    from any cell, and in each step you can move to any of the neighbor cells,
+    in other words, you can go to cells which share an edge with you current
+    cell.
+    Please note that a path of length k means visiting exactly k cells (not
+    necessarily distinct).
+    You CANNOT go off the grid.
+    A path A (of length k) is considered less than a path B (of length k) if
+    after making the ordered lists of the values on the cells that A and B go
+    through (let's call them lst_A and lst_B), lst_A is lexicographically less
+    than lst_B, in other words, there exist an integer index i (1 <= i <= k)
+    such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
+    lst_A[j] = lst_B[j].
+    It is guaranteed that the answer is unique.
+    Return an ordered list of the values on the cells that the minimum path go through.
+
+    Examples:
+
+        Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
+        Output: [1, 2, 1]
+
+        Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
+        Output: [1]
+    """
+    n = len(grid)
+    
+    # Find position of value 1 (the minimum value)
+    pos_1 = None
+    for i in range(n):
+        for j in range(n):
+            if grid[i][j] == 1:
+                pos_1 = (i, j)
+                break
+        if pos_1:
+            break
+    
+    # Find the minimum neighbor of cell containing 1
+    i, j = pos_1
+    min_neighbor = float('inf')
+    directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]
+    for di, dj in directions:
+        ni, nj = i + di, j + dj
+        if 0 <= ni < n and 0 <= nj < n:
+            min_neighbor = min(min_neighbor, grid[ni][nj])
+    
+    # The minimum path starting from 1 will alternate between 1 and its minimum neighbor
+    # This is because we can always go back to 1 from any neighbor
+    result = []
+    for step in range(k):
+        if step % 2 == 0:
+            result.append(1)
+        else:
+            result.append(min_neighbor)
+    
+    return result