Overview

The problem asks to choose an integer coordinate \(T\) on a number line such that all \(N\) friends can reach it without crossing any forbidden points, minimizing the total movement cost. Since forbidden points divide the number line into “intervals,” we determine whether all friends belong to the same interval and find the optimal meeting point within that interval.

Analysis

Key Insight: Interval Division by Forbidden Points

Since the number line is one-dimensional, it is impossible to move past forbidden points. If we sort the forbidden points \(B\), the number line is divided into the following intervals:

\[(-\infty, B_0),\quad (B_0, B_1),\quad (B_1, B_2),\quad \dots,\quad (B_{M-1}, +\infty)\]

Since there must be no forbidden point between friend \(A_i\) and destination \(T\), \(T\) and all friends must belong to the same interval.

Determination Method

To determine which interval friend \(A_i\) belongs to, we can use binary search on the sorted array \(B\) to find “the number of elements in \(B\) less than \(A_i\),” which gives us the interval index. If not all friends have the same interval index, the answer is \(-1\).

Choosing the Optimal \(T\)

Once we know that all friends are in the same interval \([lo, hi]\) (integer range), we find \(T\) that minimizes \(\sum |A_i - T|\).

Without constraints, the \(T\) that minimizes \(\sum |A_i - T|\) is the median of \(A\). When there is an even number of elements, any value from \(A[N/2-1]\) to \(A[N/2]\) is optimal.

If this optimal value falls within the interval \([lo, hi]\), we use it as is; otherwise, we clamp (restrict) it to the boundary of the interval. This works because \(\sum |A_i - T|\) is a convex function that monotonically increases as \(T\) moves away from the median.

Algorithm

1. Sort \(A\) and \(B\)
2. If \(M = 0\): No constraints, so compute the cost at the median
3. If \(M \geq 1\):
   • Use binary search to check if all friends belong to the same interval (if not, output \(-1\))
   • Determine the integer range \([lo, hi]\) of the interval (if \(lo > hi\), output \(-1\))
   • Clamp the median (optimal value) to \([lo, hi]\) to determine \(T\)
   • Compute the cost \(\sum |A_i - T|\) in \(O(N)\) using prefix sums of \(A\)

Efficient Cost Calculation

If \(A\) is sorted and we have a prefix sum array \(\text{prefix}\), letting \(p\) be the number of elements \(\leq T\):

\[\text{cost} = T \cdot p - \text{prefix}[p] + (\text{prefix}[N] - \text{prefix}[p]) - T \cdot (N - p)\]

Complexity

• Time complexity: \(O(N \log N + M \log M)\) (dominated by sorting)
• Space complexity: \(O(N + M)\)

Implementation Notes

• The interval boundaries are open intervals, so as integer coordinates, the valid range is from \(B[seg-1]+1\) to \(B[seg]-1\). Note that if adjacent forbidden points are 1 apart, then \(lo > hi\), meaning no valid integer exists.

• Clamping the median: When there is an even number of elements, consider the overlap between the optimal range \([A[(N-1)/2],\ A[N/2]]\) and \([lo, hi]\).

• Duplicate values in \(A_i\) are handled without issues (they are uniformly handled through sorting and prefix sums).

  Source Code

import sys
from bisect import bisect_left, bisect_right

def solve():
    input_data = sys.stdin.buffer.read().split()
    idx = 0
    N = int(input_data[idx]); idx += 1
    M = int(input_data[idx]); idx += 1
    
    A = []
    for i in range(N):
        A.append(int(input_data[idx])); idx += 1
    
    B = []
    for i in range(M):
        B.append(int(input_data[idx])); idx += 1
    
    B.sort()
    A.sort()
    
    # All friends must be reachable from T without crossing any forbidden point.
    # This means for every friend A_i, the open interval (min(A_i, T), max(A_i, T)) 
    # must contain no forbidden point. Also T must not be forbidden.
    #
    # So T must be in the same "segment" (between consecutive forbidden points) as ALL friends.
    # 
    # The forbidden points partition the number line into segments:
    # (-inf, B[0]), (B[0], B[1]), ..., (B[M-1], +inf)
    # and also single points that are forbidden are excluded.
    #
    # All friends must lie in the same segment, and T must also lie in that segment.
    # If friends are in different segments, answer is -1.
    
    if M == 0:
        # No forbidden points. T can be any integer. Optimal is median of A.
        median = A[N // 2] if N % 2 == 1 else A[N // 2]
        # Actually for even N, any value between A[N//2-1] and A[N//2] works.
        # We just pick median = A[(N-1)//2] or A[N//2], both give same cost.
        # Let's compute cost at median
        # For minimum sum of absolute deviations, pick median
        # With even N, any point in [A[N//2-1], A[N//2]] is optimal
        median = A[N // 2]  # This works for both odd and even
        total = sum(abs(a - median) for a in A)
        print(total)
        return
    
    # Find which segment each friend is in.
    # Segment for a point x (not in B): 
    #   segment index = number of B values < x = bisect_left(B, x)
    # All friends must have the same segment index.
    
    seg = bisect_left(B, A[0])
    for i in range(1, N):
        if bisect_left(B, A[i]) != seg:
            print(-1)
            return
    
    # All friends are in the same segment.
    # Determine the range of valid T values in this segment.
    # Segment `seg` means: 
    #   if seg == 0: T in (-inf, B[0])
    #   if seg == M: T in (B[M-1], +inf)
    #   else: T in (B[seg-1], B[seg])
    
    # T must be integer and not forbidden (which is guaranteed if in open interval between consecutive B's)
    
    if seg == 0:
        lo = -10**18
        hi = B[0] - 1
    elif seg == M:
        lo = B[M - 1] + 1
        hi = 10**18
    else:
        lo = B[seg - 1] + 1
        hi = B[seg] - 1
    
    if lo > hi:
        # No valid integer in this segment
        print(-1)
        return
    
    # Now find optimal T in [lo, hi] that minimizes sum |A_i - T|.
    # Optimal unconstrained T is the median of A.
    # Since A is sorted, median index is (N-1)//2 for lower median.
    # For even N, any T in [A[N//2-1], A[N//2]] gives minimum.
    # We clamp to [lo, hi].
    
    # Lower and upper median
    lower_med = A[(N - 1) // 2]
    upper_med = A[N // 2]
    
    # The optimal unconstrained range is [lower_med, upper_med].
    # Clamp this range to [lo, hi].
    
    # If the ranges don't overlap, pick the closest endpoint.
    if upper_med < lo:
        T = lo
    elif lower_med > hi:
        T = hi
    else:
        # Overlap: pick any point in the overlap, e.g., clamp lower_med to [lo, hi]
        T = max(lower_med, lo)
        T = min(T, hi)
    
    # Compute cost using prefix sums since A is sorted
    # sum |A_i - T| = T * count_left - sum_left + sum_right - T * count_right
    # where left = A_i <= T, right = A_i > T
    
    prefix = [0] * (N + 1)
    for i in range(N):
        prefix[i + 1] = prefix[i] + A[i]
    
    pos = bisect_right(A, T)  # number of elements <= T
    sum_left = prefix[pos]
    sum_right = prefix[N] - prefix[pos]
    cost = T * pos - sum_left + sum_right - T * (N - pos)
    
    print(cost)

solve()

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