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2194. Cells in a Range on an Excel Sheet

Description

A cell (r, c) of an excel sheet is represented as a string "<col><row>" where:

  • <col> denotes the column number c of the cell. It is represented by alphabetical letters.
    • For example, the 1st column is denoted by 'A', the 2nd by 'B', the 3rd by 'C', and so on.
  • <row> is the row number r of the cell. The rth row is represented by the integer r.

You are given a string s in the format "<col1><row1>:<col2><row2>", where <col1> represents the column c1, <row1> represents the row r1, <col2> represents the column c2, and <row2> represents the row r2, such that r1 <= r2 and c1 <= c2.

Return the list of cells (x, y) such that r1 <= x <= r2 and c1 <= y <= c2. The cells should be represented as strings in the format mentioned above and be sorted in non-decreasing order first by columns and then by rows.

 

Example 1:

Input: s = "K1:L2"
Output: ["K1","K2","L1","L2"]
Explanation:
The above diagram shows the cells which should be present in the list.
The red arrows denote the order in which the cells should be presented.

Example 2:

Input: s = "A1:F1"
Output: ["A1","B1","C1","D1","E1","F1"]
Explanation:
The above diagram shows the cells which should be present in the list.
The red arrow denotes the order in which the cells should be presented.

 

Constraints:

  • s.length == 5
  • 'A' <= s[0] <= s[3] <= 'Z'
  • '1' <= s[1] <= s[4] <= '9'
  • s consists of uppercase English letters, digits and ':'.

Solutions

Solution 1: Simulation

We directly traverse all the cells within the range and add them to the answer array.

The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$, where $m$ and $n$ are the range of rows and columns, respectively.

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class Solution:
    def cellsInRange(self, s: str) -> List[str]:
        return [
            chr(i) + str(j)
            for i in range(ord(s[0]), ord(s[-2]) + 1)
            for j in range(int(s[1]), int(s[-1]) + 1)
        ]

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class Solution {
    public List<String> cellsInRange(String s) {
        List<String> ans = new ArrayList<>();
        for (char i = s.charAt(0); i <= s.charAt(3); ++i) {
            for (char j = s.charAt(1); j <= s.charAt(4); ++j) {
                ans.add(i + "" + j);
            }
        }
        return ans;
    }
}

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class Solution {
public:
    vector<string> cellsInRange(string s) {
        vector<string> ans;
        for (char i = s[0]; i <= s[3]; ++i) {
            for (char j = s[1]; j <= s[4]; ++j) {
                ans.push_back({i, j});
            }
        }
        return ans;
    }
};

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func cellsInRange(s string) (ans []string) {
    for i := s[0]; i <= s[3]; i++ {
        for j := s[1]; j <= s[4]; j++ {
            ans = append(ans, string(i)+string(j))
        }
    }
    return
}

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function cellsInRange(s: string): string[] {
    const ans: string[] = [];
    for (let i = s.charCodeAt(0); i <= s.charCodeAt(3); ++i) {
        for (let j = s.charCodeAt(1); j <= s.charCodeAt(4); ++j) {
            ans.push(String.fromCharCode(i) + String.fromCharCode(j));
        }
    }
    return ans;
}

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