862. Find And Replace in String

Problem

To some string S, we will perform some replacement operations that replace groups of letters with new ones (not necessarily the same size).

Each replacement operation has 3 parameters: a starting index i, a source word x and a target word y.  The rule is that if x starts at position i in the original string S, then we will replace that occurrence of x with y.  If not, we do nothing.

For example, if we have S = "abcd" and we have some replacement operation i = 2, x = "cd", y = "ffff", then because "cd" starts at position 2 in the original string S, we will replace it with "ffff".

Using another example on S = "abcd", if we have both the replacement operation i = 0, x = "ab", y = "eee", as well as another replacement operation i = 2, x = "ec", y = "ffff", this second operation does nothing because in the original string S[2] = 'c', which doesn't match x[0] = 'e'.

All these operations occur simultaneously.  It's guaranteed that there won't be any overlap in replacement: for example, S = "abc", indexes = [0, 1], sources = ["ab","bc"] is not a valid test case.

Example 1:

Input: S = "abcd", indexes = [0,2], sources = ["a","cd"], targets = ["eee","ffff"]
Output: "eeebffff"
Explanation: "a" starts at index 0 in S, so it's replaced by "eee".
"cd" starts at index 2 in S, so it's replaced by "ffff".

Example 2:

Input: S = "abcd", indexes = [0,2], sources = ["ab","ec"], targets = ["eee","ffff"]
Output: "eeecd"
Explanation: "ab" starts at index 0 in S, so it's replaced by "eee". 
"ec" doesn't starts at index 2 in the original S, so we do nothing.

Notes:

• 0 <=&nbsp;indexes.length =&nbsp;sources.length =&nbsp;targets.length <= 100
• 0&nbsp;<&nbsp;indexes[i]&nbsp;< S.length <= 1000
• All characters in given inputs are lowercase letters.

Solution 1

/**
 * @param {string} S
 * @param {number[]} indexes
 * @param {string[]} sources
 * @param {string[]} targets
 * @return {string}
 */
var findReplaceString = function(S, indexes, sources, targets) {
  var len = S.length;
  var len2 = indexes.length;
  var map = {};
  var res = '';
  var i = 0;

  if (len2 === 0) return S;

  for (var k = 0; k < len2; k++) {
    map[indexes[k]] = [sources[k], targets[k]];
  }

  while (i < len) {
    if (map[i] && S.substr(i, map[i][0].length) === map[i][0]) {
      res += map[i][1];
      i += Math.max(map[i][0].length, 1);
    } else {
      res += S[i];
      i += 1;
    }
  }

  return res;
};

Explain:

nope.

Complexity:

• Time complexity : O(n). n 为 S.length。
• Space complexity : O(n).

Solution 2

/**
 * @param {string} S
 * @param {number[]} indexes
 * @param {string[]} sources
 * @param {string[]} targets
 * @return {string}
 */
var findReplaceString = function(S, indexes, sources, targets) {
  var len = indexes.length;
  var sorted = [];
  var map = {};
  var index = 0;

  if (len === 0) return S;

  for (var i = 0; i < len; i++) {
    map[indexes[i]] = i;
    sorted.push(indexes[i]);
  }

  sorted.sort((a, b) => a - b);

  for (var j = len - 1; j >= 0; j--) {
    index = map[sorted[j]];
    if (S.substr(sorted[j], sources[index].length) === sources[index]) {
      S = S.substr(0, sorted[j]) + targets[index] + S.substr(sorted[j] + sources[index].length);
    }
  }

  return S;
};

Explain:

给 indexes 排序，然后从后往前依次替换。

Complexity:

• Time complexity : O(n * log(n)). n 为 indexes.length。
• Space complexity : O(n).