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844. Backspace String Compare

Description

Given two strings s and t, return true if they are equal when both are typed into empty text editors. '#' means a backspace character.

Note that after backspacing an empty text, the text will continue empty.

 

Example 1:

Input: s = "ab#c", t = "ad#c"
Output: true
Explanation: Both s and t become "ac".

Example 2:

Input: s = "ab##", t = "c#d#"
Output: true
Explanation: Both s and t become "".

Example 3:

Input: s = "a#c", t = "b"
Output: false
Explanation: s becomes "c" while t becomes "b".

 

Constraints:

  • 1 <= s.length, t.length <= 200
  • s and t only contain lowercase letters and '#' characters.

 

Follow up: Can you solve it in O(n) time and O(1) space?

Solutions

Solution 1

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class Solution:
    def backspaceCompare(self, s: str, t: str) -> bool:
        i, j, skip1, skip2 = len(s) - 1, len(t) - 1, 0, 0
        while i >= 0 or j >= 0:
            while i >= 0:
                if s[i] == '#':
                    skip1 += 1
                    i -= 1
                elif skip1:
                    skip1 -= 1
                    i -= 1
                else:
                    break
            while j >= 0:
                if t[j] == '#':
                    skip2 += 1
                    j -= 1
                elif skip2:
                    skip2 -= 1
                    j -= 1
                else:
                    break
            if i >= 0 and j >= 0:
                if s[i] != t[j]:
                    return False
            elif i >= 0 or j >= 0:
                return False
            i, j = i - 1, j - 1
        return True
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class Solution {
    public boolean backspaceCompare(String s, String t) {
        int i = s.length() - 1, j = t.length() - 1;
        int skip1 = 0, skip2 = 0;
        for (; i >= 0 || j >= 0; --i, --j) {
            while (i >= 0) {
                if (s.charAt(i) == '#') {
                    ++skip1;
                    --i;
                } else if (skip1 > 0) {
                    --skip1;
                    --i;
                } else {
                    break;
                }
            }
            while (j >= 0) {
                if (t.charAt(j) == '#') {
                    ++skip2;
                    --j;
                } else if (skip2 > 0) {
                    --skip2;
                    --j;
                } else {
                    break;
                }
            }
            if (i >= 0 && j >= 0) {
                if (s.charAt(i) != t.charAt(j)) {
                    return false;
                }
            } else if (i >= 0 || j >= 0) {
                return false;
            }
        }
        return true;
    }
}
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class Solution {
public:
    bool backspaceCompare(string s, string t) {
        int i = s.size() - 1, j = t.size() - 1;
        int skip1 = 0, skip2 = 0;
        for (; i >= 0 || j >= 0; --i, --j) {
            while (i >= 0) {
                if (s[i] == '#') {
                    ++skip1;
                    --i;
                } else if (skip1) {
                    --skip1;
                    --i;
                } else
                    break;
            }
            while (j >= 0) {
                if (t[j] == '#') {
                    ++skip2;
                    --j;
                } else if (skip2) {
                    --skip2;
                    --j;
                } else
                    break;
            }
            if (i >= 0 && j >= 0) {
                if (s[i] != t[j]) return false;
            } else if (i >= 0 || j >= 0)
                return false;
        }
        return true;
    }
};
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func backspaceCompare(s string, t string) bool {
    i, j := len(s)-1, len(t)-1
    skip1, skip2 := 0, 0
    for ; i >= 0 || j >= 0; i, j = i-1, j-1 {
        for i >= 0 {
            if s[i] == '#' {
                skip1++
                i--
            } else if skip1 > 0 {
                skip1--
                i--
            } else {
                break
            }
        }
        for j >= 0 {
            if t[j] == '#' {
                skip2++
                j--
            } else if skip2 > 0 {
                skip2--
                j--
            } else {
                break
            }
        }
        if i >= 0 && j >= 0 {
            if s[i] != t[j] {
                return false
            }
        } else if i >= 0 || j >= 0 {
            return false
        }
    }
    return true
}
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function backspaceCompare(s: string, t: string): boolean {
    let i = s.length - 1;
    let j = t.length - 1;
    while (i >= 0 || j >= 0) {
        let skip = 0;
        while (i >= 0) {
            if (s[i] === '#') {
                skip++;
            } else if (skip !== 0) {
                skip--;
            } else {
                break;
            }
            i--;
        }
        skip = 0;
        while (j >= 0) {
            if (t[j] === '#') {
                skip++;
            } else if (skip !== 0) {
                skip--;
            } else {
                break;
            }
            j--;
        }
        if (s[i] !== t[j]) {
            return false;
        }
        i--;
        j--;
    }
    return true;
}
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impl Solution {
    pub fn backspace_compare(s: String, t: String) -> bool {
        let (s, t) = (s.as_bytes(), t.as_bytes());
        let (mut i, mut j) = (s.len(), t.len());
        while i != 0 || j != 0 {
            let mut skip = 0;
            while i != 0 {
                if s[i - 1] == b'#' {
                    skip += 1;
                } else if skip != 0 {
                    skip -= 1;
                } else {
                    break;
                }
                i -= 1;
            }
            skip = 0;
            while j != 0 {
                if t[j - 1] == b'#' {
                    skip += 1;
                } else if skip != 0 {
                    skip -= 1;
                } else {
                    break;
                }
                j -= 1;
            }
            if i == 0 && j == 0 {
                break;
            }
            if i == 0 || j == 0 {
                return false;
            }
            if s[i - 1] != t[j - 1] {
                return false;
            }
            i -= 1;
            j -= 1;
        }
        true
    }
}

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