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@ -6,39 +6,95 @@ const SEPARATOR = '$';
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* @return {number[]}
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*/
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function buildZArray(zString) {
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const zArray = new Array(zString.length);
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// Initiate zArray and fill it with zeros.
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const zArray = new Array(zString.length).fill(null).map(() => 0);
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let left = 0;
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let right = 0;
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let k = 0;
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// Z box boundaries.
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let zBoxLeftIndex = 0;
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let zBoxRightIndex = 0;
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for (let i = 1; i < zString.length; i += 1) {
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if (i > right) {
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left = i;
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right = i;
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// Position of current zBox character that is also a position of
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// the same character in prefix.
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// For example:
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// Z string: ab$xxabxx
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// Indices: 012345678
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// Prefix: ab.......
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// Z box: .....ab..
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// Z box shift for 'a' would be 0 (0-position in prefix and 0-position in Z box)
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// Z box shift for 'b' would be 1 (1-position in prefix and 1-position in Z box)
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let zBoxShift = 0;
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while (right < zString.length && zString[right - left] === zString[right]) {
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right += 1;
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// Go through all characters of the zString.
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for (let charIndex = 1; charIndex < zString.length; charIndex += 1) {
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if (charIndex > zBoxRightIndex) {
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// We're OUTSIDE of Z box. In other words this is a case when we're
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// starting from Z box of size 1.
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// In this case let's make current character to be a Z box of length 1.
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zBoxLeftIndex = charIndex;
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zBoxRightIndex = charIndex;
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// Now let's go and check current and the following characters to see if
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// they are the same as a prefix. By doing this we will also expand our
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// Z box. For example if starting from current position we will find 3
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// more characters that are equal to the ones in the prefix we will expand
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// right Z box boundary by 3.
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while (
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zBoxRightIndex < zString.length &&
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zString[zBoxRightIndex - zBoxLeftIndex] === zString[zBoxRightIndex]
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) {
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// Expanding Z box right boundary.
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zBoxRightIndex += 1;
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}
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zArray[i] = right - left;
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right -= 1;
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// Now we may calculate how many characters starting from current position
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// are are the same as the prefix. We may calculate it by difference between
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// right and left Z box boundaries.
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zArray[charIndex] = zBoxRightIndex - zBoxLeftIndex;
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// Move right Z box boundary left by one position just because we've used
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// [zBoxRightIndex - zBoxLeftIndex] index calculation above.
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zBoxRightIndex -= 1;
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} else {
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k = i - left;
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if (zArray[k] < (right - i) + 1) {
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zArray[i] = zArray[k];
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// We're INSIDE of Z box.
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// Calculate corresponding Z box shift. Because we want to copy the values
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// from zArray that have been calculated before.
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zBoxShift = charIndex - zBoxLeftIndex;
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// Check if the value that has been already calculated before
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// leaves us inside of Z box or it goes beyond the checkbox
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// right boundary.
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if (zArray[zBoxShift] < (zBoxRightIndex - charIndex) + 1) {
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// If calculated value don't force us to go outside Z box
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// then we're safe and we may simply use previously calculated value.
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zArray[charIndex] = zArray[zBoxShift];
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} else {
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left = i;
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while (right < zString.length && zString[right - left] === zString[right]) {
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right += 1;
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// In case if previously calculated values forces us to go outside of Z box
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// we can't safely copy previously calculated zArray value. It is because
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// we are sure that there is no further prefix matches outside of Z box.
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// Thus such values must be re-calculated and reduced to certain point.
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// To do so we need to shift left boundary of Z box to current position.
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zBoxLeftIndex = charIndex;
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// And start comparing characters one by one as we normally do for the case
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// when we are outside of checkbox.
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while (
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zBoxRightIndex < zString.length &&
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zString[zBoxRightIndex - zBoxLeftIndex] === zString[zBoxRightIndex]
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) {
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zBoxRightIndex += 1;
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}
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zArray[i] = right - left;
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right -= 1;
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zArray[charIndex] = zBoxRightIndex - zBoxLeftIndex;
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zBoxRightIndex -= 1;
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}
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}
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}
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// Return generated zArray.
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return zArray;
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}
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