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Leetcode #1198: Find Smallest Common Element in All Rows

In this guide, we solve Leetcode #1198 Find Smallest Common Element in All Rows in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

Given an m x n matrix mat where every row is sorted in strictly increasing order, return the smallest common element in all rows. If there is no common element, return -1.

Quick Facts

  • Difficulty: Medium
  • Premium: Yes
  • Tags: Array, Hash Table, Binary Search, Counting, Matrix

Intuition

Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.

By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.

Approach

Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.

This keeps the solution linear while remaining easy to explain in an interview setting.

Steps:

  • Initialize a hash map for seen items or counts.
  • Iterate through the input, querying/updating the map.
  • Return the first valid result or the final computed value.

Example

Input: mat = [[1,2,3,4,5],[2,4,5,8,10],[3,5,7,9,11],[1,3,5,7,9]] Output: 5

Python Solution

class Solution: def smallestCommonElement(self, mat: List[List[int]]) -> int: cnt = Counter() for row in mat: for x in row: cnt[x] += 1 if cnt[x] == len(mat): return x return -1

Complexity

The time complexity is O(m×n)O(m \times n)O(m×n), and the space complexity is O(104)O(10^4)O(104). The space complexity is O(104)O(10^4)O(104).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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