Algorithm
Problem Name: Data Structures -
In this HackerRank in Data Structures -
The square-ten tree decomposition of an array is defined as follows:
- The lowest (0**th) level of the square-ten tree consists of single array elements in their natural order.
- The k**th level (starting from 1 ) of the square-ten tree consists of subsequent array subsegments of length 10**2k-1 in their natural order. Thus, the 1st level contains subsegments of length 10**2**1-1 = 10 the 2nd level contains subsegments of length 10**2**2-i = 100 the 3rd level contains subsegments of length 10**2**3-1 = 10000 etc.
The image below depicts the bottom-left corner (i.e., the first 128 array elements) of the table representing a square-ten tree. The levels are numbered from bottom to top:
Input Format
The first line contains a single integer denoting L.
The second line contains a single integer denoting R .
Constraints
- 1 <= L <= R <= 10**10**6
- The numbers in input do not contain leading zeroes.
Sample Input 0
1
10
Sample Output 0
1
1 1
Explanation 0
Segment [1,10] belongs to level 1 of the square-ten tree.
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define DIGIT(a) ((a) - '0')
#define CHAR(a) ((a) + '0')
char L[1000002];
char R[1000002];
char amount[1000002];
int main()
{
int lvls[42];
char * amts[42] = {0};
int lenL, lenR, diff1, left, right, pos, groupsz, level, max, rmd, carry, i, n = 0;
scanf("%s", L);
scanf("%s", R);
lenL = strlen(L);
lenR = strlen(R);
if (lenL < lenR)
{
memmove(L + lenR - lenL, L, lenL + 1);
memset(L, '0', lenR - lenL);
lenL = lenR;
}
for (i = 0; i < lenL; ++i)
if (L[i] != R[i]) break;
diff1 = i;
if (diff1 >= lenL-1)
{
printf("1\n0 %d\n", 1 + R[lenR-1] - L[lenL-1]);
return 0;
}
level = 0;
if (L[lenL-1] == '0')
{
L[lenL-1] = '1';
lvls[n] = level;
amts[n] = malloc(1 + 1);
if (amts[n] == NULL)
return 1;
strcpy(amts[n], "1");
n++;
}
else if (L[lenL-1] != '1')
{
lvls[n] = level;
amts[n] = malloc(1 + 1);
if (amts[n] == NULL)
return 1;
amts[n][0] = CHAR(11 - DIGIT(L[lenL-1]));
amts[n][1] = '\0';
n++;
for (pos = lenL-2; L[pos] == '9'; --pos)
;
L[pos] += 1;
for (i = pos+1; i < lenL; ++i)
L[i] = '0';
}
groupsz = 1;
left = right = lenL-2;
while (diff1 < left)
{
level += 1;
rmd = 0;
max = '0';
for (i = right; i >= left; --i)
{
if (L[i] > max) max = L[i];
rmd -= DIGIT(L[i]);
if (rmd < 0)
{
amount[i] = CHAR(rmd + 10);
rmd = -1;
}
else
{
amount[i] = CHAR(rmd);
rmd = 0;
}
}
if (max > '0')
{
for (pos = left-1; L[pos] == '9'; --pos)
;
L[pos] += 1;
for (i = pos+1; i < = right; ++i)
L[i] = '0';
lvls[n] = level;
amts[n] = malloc(groupsz + 1);
if (amts[n] == NULL)
return 1;
memcpy(amts[n], &amount[left], groupsz);
amts[n][groupsz] = '\0';
n++;
}
right -= groupsz;
groupsz *= 2;
left = MAX(0, right - groupsz + 1);
}
/* take truncated/diff group */
level += 1;
rmd = 0;
max = '0';
for (i = right; i >= left; --i)
{
rmd += R[i] - L[i];
if (rmd < 0)
{
amount[i] = CHAR(rmd + 10);
rmd = -1;
}
else
{
amount[i] = CHAR(rmd);
rmd = 0;
}
if (amount[i] > max) max = amount[i];
}
if (max > '0')
{
lvls[n] = level;
amts[n] = malloc(right + 1 - left + 1);
if (amts[n] == NULL)
return 1;
memcpy(amts[n], &amount[left], right + 1 - left);
amts[n][right+1-left] = '\0';
n++;
}
left = right + 1;
groupsz /= 2;
right = left + groupsz - 1;
while (left < lenR-1)
{
level -= 1;
max = '0';
for (i = left; i < = right; ++i)
if (R[i] > '0') max = R[i];
if (max > '0')
{
if (lvls[n-1] == level)
{
/* add amount to amts[n-1] */
carry = 0;
for (i = right; i >= left; --i)
{
carry += DIGIT(R[i]) + DIGIT(amts[n-1][i-left]);
if (carry >= 10)
{
amount[i] = CHAR(carry - 10);
carry = 1;
}
else
{
amount[i] = CHAR(carry);
carry = 0;
}
}
if (carry)
{
amount[i] = '1';
free(amts[n-1]);
amts[n-1] = malloc(1 + groupsz + 1);
if (amts[n-1] == NULL)
return 1;
memcpy(amts[n-1], &amount[i], 1 + groupsz);
amts[n-1][1+groupsz] = '\0';
}
else
{
memcpy(amts[n-1], &amount[left], groupsz);
}
}
else
{
lvls[n] = level;
amts[n] = malloc(groupsz + 1);
if (amts[n] == NULL)
return 1;
memcpy(amts[n], &R[left], groupsz);
amts[n][groupsz] = '\0';
n++;
}
}
left = right + 1;
groupsz /= 2;
right = left + groupsz - 1;
}
if (R[lenR-1] > '0')
{
level = 0;
if (lvls[n-1] == level)
{
/* add amount to amts[n-1] */
carry = DIGIT(R[lenR-1]) + DIGIT(amts[n-1][0]);
if (carry >= 10)
{
free(amts[n-1]);
amts[n-1] = malloc(2 + 1);
if (amts[n-1] == NULL)
return 1;
sprintf(amts[n-1], "%d", carry);
}
else
{
amts[n-1][0] = CHAR(carry);
}
}
else
{
lvls[n] = level;
amts[n] = malloc(1 + 1);
if (amts[n] == NULL)
return 1;
amts[n][0] = R[lenR-1];
amts[n][1] = '\0';
n++;
}
}
printf("%d\n", n);
for (i = 0; i < n; ++i)
{
for (left = 0; amts[i][left] == '0'; ++left)
;
printf("%d %s\n", lvls[i], &amts[i][left]);
}
return 0;
}
#2 Code Example with C++ Programming
Code -
C++ Programming
#include <cmath>
#include <cstdio>
#include < vector>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stdlib.h>
#include <sstream>
using namespace std;
struct Block {
unsigned int level;
string count;
};
void pushBlocks(vector < Block> &blocks, unsigned level, const string& s) {
auto it = s.rbegin();
while (*it == '0') it++;
if (it != s.rend()) blocks.push_back({level, string(it, s.rend())});
}
int main() {
string l, r;
vector < Block> blocks;
cin >> l >> r;
if (l == r) {
blocks.push_back({ 0, "1" });
}
else {
// skip matching start
if (l.length() == r.length()) {
int i = 0;
while (l[i] == r[i]) i++;
l = l.substr(i);
r = r.substr(i);
}
// reverse strings for easier indexing from lowest digit
reverse(l.begin(), l.end());
reverse(r.begin(), r.end());
const int llen = l.length();
const int rlen = r.length();
int blockEnd = 1;
unsigned int level = 0;
int carry = -1;
stringstream output;
// blocks needed raise 'l' to next blockEnd
int i = 0;
for (; i < rlen; i++) {
if (i >= blockEnd) {
pushBlocks(blocks, level, output.str());
level++;
blockEnd *= 2;
output.str("");
}
if (blockEnd >= rlen) {
break;
}
int x = i < llen ? l[i] - '0' : 0;
int diff = (10 - x - carry) % 10;
carry = x + carry + diff >= 10;
output << (char)('0' + diff);
}
// top block (simply 'r' - 'l')
for (int j = i; j < rlen; j++) {
int x = j < llen ? l[j] - '0' : 0;
int y = r[j] - '0';
int diff = (10 + y - x - carry) % 10;
carry = y - x - carry < 0;
output << (char)('0' + diff);
}
pushBlocks(blocks, level, output.str());
level--;
blockEnd /= 2;
output.str("");
// add remaining 'r' blocks
while (blockEnd > 1) {
pushBlocks(blocks, level, r.substr(blockEnd / 2, blockEnd / 2));
level--;
blockEnd /= 2;
}
pushBlocks(blocks, level, r.substr(0, 1));
}
// print result
cout << blocks.size() << endl;
for (auto iter = blocks.begin(); iter != blocks.end(); iter++) {
cout << iter->level << " " << iter->count << endl;
}
return 0;
}
#3 Code Example with Java Programming
Code -
Java Programming
import java.util.Scanner;
import java.util.Arrays;
// I recommend skipping this problem. The problem statement is way too convoluted.
// However, here are some takeaway concepts from this problem:
// - Implementing a custom BigInt as a byte[]
// - Implementing log2(int n)
// - Realizing that our algorithm can process the giant numbers in portions
// - Runtime: O(n + m) where n = # of digits in L, m = # of digits in R (for interval [L,R]).
// - Numbers can literally have millions of digits in this problem. An "int" or "long" is not big enough to store these numbers. Although Java's BigInteger is big enough, it turns out to be too slow for this problem. I wrote a custom "BigInt" class to speed up calculations.
// - To achieve linear runtime, we need an algorithm that splits up these giant numbers into portions and processes them separately. A great way to do this is to split by level, as done below.
// - This was a very difficult problem. You must have both linear runtime and efficient code to pass all testcases.
public class Solution {
public static void main(String[] args) {
/* Read and save input */
Scanner scan = new Scanner(System.in);
String strL = new BigInt(scan.next()).subtract(BigInt.ONE).toString(); // subtract 1 since it's [L,R] inclusive
String strR = scan.next();
scan.close();
/* Calculate interval sizes (by just saving # of digits) */
int [] intervalDigits = new int[log2(strR.length()) + 3]; // The +3 gives us an estimate of the size we need
for (int k = 0; k < intervalDigits.length; k++) {
intervalDigits[k] = digitsInInterval(k);
}
/* Initialize variables */
StringBuilder sb = new StringBuilder();
int endL = strL.length();
int endR = strR.length();
BigInt upperBound = BigInt.ONE;
boolean carry = false;
boolean lastIteration = false;
int blockCount = 0;
int level = 0;
/* Calculate counts for increasing segment sizes */
while (!lastIteration) {
/* Get portion of each String corresponding to current level */
int numDigits = intervalDigits[level + 1] - intervalDigits[level];
int startL = Math.max(endL - numDigits, 0);
int startR = Math.max(endR - numDigits, 0);
BigInt numL = (endL == 0) ? BigInt.ZERO : new BigInt(strL.substring(startL, endL));
if (carry) {
numL = numL.add(BigInt.ONE);
}
/* Calculate upper bound */
if (startR == 0) {
upperBound = new BigInt(strR.substring(startR, endR));
lastIteration = true;
} else {
upperBound = BigInt.tenToPower(numDigits);
}
/* If not skipping this level, process it */
if ((!numL.equals(BigInt.ZERO) && !numL.equals(upperBound)) || startR == 0) {
BigInt count = upperBound.subtract(numL);
carry = true;
blockCount++;
sb.append(level + " " + count + "\n");
}
/* Update variables for next iteration */
endL = startL;
endR = startR;
level++;
}
StringBuilder sb2 = new StringBuilder();
level = 0;
endR = strR.length();
/* Calculate counts for decreasing segment sizes */
while (true) {
/* Calculate number of nodes in current level */
int numDigits = intervalDigits[level + 1] - intervalDigits[level];
int startR = Math.max(endR - numDigits, 0);
if (startR == 0) {
break;
}
BigInt count = new BigInt(strR.substring(startR, endR));
/* If not skipping this level, process it */
if (!count.equals(BigInt.ZERO)) {
blockCount++;
sb2.insert(0, level + " " + count + "\n");
}
/* Update variables for next iteration */
endR = startR;
level++;
}
System.out.println(blockCount + "\n" + sb + sb2);
}
static int log2(int n) { // assumes positive number
return 31 - Integer.numberOfLeadingZeros(n);
}
static int digitsInInterval(int k) {
if (k == 0) {
return 1;
} else {
return (int) (Math.pow(2, k - 1) + 1);
}
}
}
// - Java's BigInteger is not fast enough to pass the testcases. The BigInt I create below is more efficient for this problem.
// - This link has good implementation ideas (Though they store numbers in reverse order):
// http://iwillgetthatjobatgoogle.tumblr.com/post/32583376161/writing-biginteger-better-than-jdk-one
// - BigInt numbers may be stored with leading 0s
// - BigInt only works with non-negative integers
class BigInt {
public static final BigInt ZERO = new BigInt("0");
public static final BigInt ONE = new BigInt("1");
public final byte[] digits; // will use 8 bits per digit for simplicity, even though 4 bits is enough
/* Constructor */
public BigInt(String str) {
digits = new byte[str.length()];
for (int i = 0; i < digits.length; i++) {
digits[i] = Byte.valueOf(str.substring(i, i + 1));
}
}
/* Constructor */
public BigInt(byte [] digits) {
this.digits = digits;
}
public static BigInt tenToPower(int exponent) {
byte [] digits = new byte[exponent + 1];
digits[0] = 1;
return new BigInt(digits);
}
public BigInt add(BigInt other) {
byte [] digitsA = digits;
byte [] digitsB = other.digits;
/* Create new Array to hold answer */
int newLength = Math.max(digitsA.length, digitsB.length);
if (!(digitsA[0] == 0 && digitsB[0] == 0)) {
newLength++;
}
byte [] result = new byte[newLength];
/* Do the addition */
int carry = 0;
int ptrA = digitsA.length - 1;
int ptrB = digitsB.length - 1;
int ptrR = result.length - 1;
while (ptrA >= 0 || ptrB >= 0 || carry > 0) {
int sum = carry;
if (ptrA >= 0) {
sum += digitsA[ptrA--];
}
if (ptrB >= 0) {
sum += digitsB[ptrB--];
}
result[ptrR--] = (byte) (sum % 10);
carry = sum / 10;
}
return new BigInt(result);
}
public BigInt subtract(BigInt other) { // assumes "other" is smaller than this BigInt
byte [] digitsB = other.digits;
byte [] result = Arrays.copyOf(digits, digits.length); // copy of "digitsA"
/* Do the subtraction */
int ptrB = digitsB.length - 1;
int ptrR = result.length - 1;
while (ptrB >= 0 && ptrR >= 0) {
result[ptrR] -= digitsB[ptrB];
/* if necessary, do the "borrow" */
if (result[ptrR] < 0) {
result[ptrR] += 10;
int ptrBorrow = ptrR - 1;
while (result[ptrBorrow] == 0) {
result[ptrBorrow--] = 9;
}
result[ptrBorrow]--;
}
ptrB--;
ptrR--;
}
return new BigInt(result);
}
@Override
public boolean equals(Object other) {
if (!(other instanceof BigInt)) {
return false;
}
byte [] digitsA = digits;
byte [] digitsB = ((BigInt) other).digits;
int indexA = 0;
int indexB = 0;
/* Remove leading 0s */
while (indexA < digitsA.length && digitsA[indexA] == 0) {
indexA++;
}
while (indexB < digitsB.length && digitsB[indexB] == 0) {
indexB++;
}
/* If lengths not equal, BigInts aren't equal */
int lenA = digitsA.length - indexA;
int lenB = digitsB.length - indexB;
if (lenA != lenB) {
return false;
}
/* Check to see if all digits match for the 2 BigInts */
while (indexA < digitsA.length && indexB < digitsB.length) {
if (digitsA[indexA++] != digitsB[indexB++]) {
return false;
}
}
return true;
}
@Override
public String toString() {
StringBuilder sb = new StringBuilder();
int i = 0;
/* Skip leading 0s */
while (i < digits.length && digits[i] == 0) {
i++;
}
/* Special Case: the BigInt 0 */
if (i == digits.length) {
return "0";
}
/* Create and return String */
for ( ; i < digits.length; i++) {
sb.append(digits[i]);
}
return sb.toString();
}
}
#4 Code Example with Python Programming
Code -
Python Programming
class Big:
def __init__(self, v):
self.value = self.check_zero(v)
self.size = 5000
self.size2 = 10 ** self.size
def modd(self, length, base0=False):
vv = self.check_zero(self.value[-length:])
if vv == '':
if base0:
return ''
else:
return '1' + '0' * length
return self.check_zero(self.value[-length:])
def mod_rest(self, length, is_plus):
vv = self.check_zero(self.value[-length:])
self.value = self.value[:-length]
if is_plus:
if not vv == '':
self.value = self.plus(self.value, '1')
def plus(self, value1, value2):
result = ""
size = self.size
size2 = self.size2
length = (max(len(value1), len(value2)) // size + 1) * size
value1 = '0' * (length - len(value1)) + value1
value2 = '0' * (length - len(value2)) + value2
rest = 0
for j in range(length, 0, -size):
v1 = int(value1[j - size: j])
v2 = int(value2[j - size: j])
r = v1 + v2 + rest
if r >= size2:
rest = 1
r -= size2
else:
rest = 0
result = '0' * (size - len(str(r))) + str(r) + result
if rest > 0:
result = str(rest) + result
result = self.check_zero(result)
return result
def minus(self, value1, value2):
# value1 - value2, len(value1) >= len(value2)
result = ""
size = self.size
size2 = self.size2
length = (max(len(value1), len(value2)) // size + 1) * size
value1 = '0' * (length - len(value1)) + value1
value2 = '0' * (length - len(value2)) + value2
rest = 0
for j in range(length, 0, -size):
v1 = int(value1[j - size: j])
v2 = int(value2[j - size: j])
r = v1 - v2 + rest
if r < 0:
rest = -1
r += size2
else:
rest = 0
result = '0' * (size - len(str(r))) + str(r) + result
result = self.check_zero(result)
return result
def compare(self, length):
return len(self.value) - length
def check_zero(self, k):
# if len(k) == 0:
# return k
strr = ''
for j in range(len(k)):
if k[j] == '0':
continue
else:
strr = k[j:]
break
return strr
def main(l, r):
ll = Big(l)
rr = Big(r)
indexs = [1, 1]
results1 = []
results2 = []
# print(ll.value, rr.value)
ll.value = ll.minus(ll.value, '1')
while True:
index = indexs[-2]
if ll.compare(index) <= 0 and rr.compare(index) <= 0:
results1.append(ll.minus(rr.value, ll.value))
# print(len(results1) - 1, ll.value, rr.value, results1[-1])
break
else:
l1 = ll.minus('1' + '0' * index, ll.modd(index))
# l1 = ll.plus(l1, '1')
r1 = rr.modd(index, True)
# print(len(results1), ll.value, rr.value, l1, r1, index)
ll.mod_rest(index, True)
rr.mod_rest(index, False)
results1.append(l1)
results2.append(r1)
indexs.append(indexs[-1] << 1)
# print(results1, results2)
final = []
for n, k in enumerate(results1):
strr = ll.check_zero(k)
if not strr == '':
final.append([n, strr])
for m in range(len(results2) - 1, -1, -1):
strr = ll.check_zero(results2[m])
if not strr == '':
final.append([m, strr])
print(len(final))
for ii in range(len(final)):
print(final[ii][0], final[ii][1])
if __name__ == '__main__':
main(input(), input())