Smallest number greater than Y with sum of digits equal to X
Last Updated :
10 Jun, 2021
Given two integers X and Y, find the minimal number with the sum of digits X, which is strictly greater than Y.
Examples:
Input: X = 18, Y = 99
Output: 189
Explanation:
189 is the smallest number greater than 99 having sum of digits = 18.
Input: X = 12, Y = 72
Output: 75
Explanation:
75 is the smallest number greater than 72 that has sum of digits = 12.
Naive Approach: The naive approach is iterate from Y + 1 and check if any number whose sum of digits is X or not. If we found any such number then print that number.
Time Complexity: O((R - Y)*log10N), where R is the maximum number till where we iterate and N is the number in the range [Y, R]
Auxiliary Space: O(1)
Efficient Approach: The idea is to iterate through the digits of Y from right to left, and try to increase the current digit and change the digits to the right in order to make the sum of digits equal to X. Below are the steps:
- If we are considering the (k + 1)th digit from the right and increasing it, then it is possible to make the sum of k least significant digits to be any number in the range [0, 9k].
- When such a position is found, then stop the process and print the number at that iteration.
- If k least significant digits have sum M (where 0 ? M ? 9k), then obtain the answer greedily:
- Traverse from the right to the left and insert 9 and subtract 9 from the sum of digits.
- Once, the sum is less than 9, place the remaining sum.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the minimum string
// of length d having the sum of digits s
string helper(int d, int s)
{
// Return a string of length d
string ans(d, '0');
for (int i = d - 1; i >= 0; i--) {
// Greedily put 9's in the end
if (s >= 9) {
ans[i] = '9';
s -= 9;
}
// Put remaining sum
else {
char c = (char)s + '0';
ans[i] = c;
s = 0;
}
}
return ans;
}
// Function to find the smallest
// number greater than Y
// whose sum of digits is X
string findMin(int x, int Y)
{
// Convert number y to string
string y = to_string(Y);
int n = y.size();
vector<int> p(n);
// Maintain prefix sum of digits
for (int i = 0; i < n; i++) {
p[i] = y[i] - '0';
if (i > 0)
p[i] += p[i - 1];
}
// Iterate over Y from the back where
// k is current length of suffix
for (int i = n - 1, k = 0;; i--, k++) {
// Stores current digit
int d = 0;
if (i >= 0)
d = y[i] - '0';
// Increase current digit
for (int j = d + 1; j <= 9; j++) {
// Sum upto current prefix
int r = (i > 0) * p[i - 1] + j;
// Return answer if remaining
// sum can be obtained in suffix
if (x - r >= 0 and x - r <= 9 * k) {
// Find suffix of length k
// having sum of digits x-r
string suf = helper(k, x - r);
string pre = "";
if (i > 0)
pre = y.substr(0, i);
// Append current character
char cur = (char)j + '0';
pre += cur;
// Return the result
return pre + suf;
}
}
}
}
// Driver Code
int main()
{
// Given Number and Sum
int x = 18;
int y = 99;
// Function Call
cout << findMin(x, y) << endl;
return 0;
}
// Java program for the above approach
import java.util.*;
@SuppressWarnings("unchecked")
class GFG{
// Function to return the minimum String
// of length d having the sum of digits s
static String helper(int d, int s)
{
// Return a String of length d
StringBuilder ans = new StringBuilder();
for(int i = 0; i < d; i++)
{
ans.append("0");
}
for(int i = d - 1; i >= 0; i--)
{
// Greedily put 9's in the end
if (s >= 9)
{
ans.setCharAt(i,'9');
s -= 9;
}
// Put remaining sum
else
{
char c = (char)(s + (int)'0');
ans.setCharAt(i, c);
s = 0;
}
}
return ans.toString();
}
// Function to find the smallest
// number greater than Y
// whose sum of digits is X
static String findMin(int x, int Y)
{
// Convert number y to String
String y = Integer.toString(Y);
int n = y.length();
ArrayList p = new ArrayList();
for(int i = 0; i < n; i++)
{
p.add(0);
}
// Maintain prefix sum of digits
for(int i = 0; i < n; i++)
{
p.add(i, (int)((int) y.charAt(i) - (int)'0'));
if (i > 0)
{
p.add(i, (int)p.get(i) +
(int)p.get(i - 1));
}
}
// Iterate over Y from the back where
// k is current length of suffix
for(int i = n - 1, k = 0;; i--, k++)
{
// Stores current digit
int d = 0;
if (i >= 0)
{
d = (int) y.charAt(i) - (int)'0';
}
// Increase current digit
for(int j = d + 1; j <= 9; j++)
{
int r = j;
// Sum upto current prefix
if (i > 0)
{
r += (int) p.get(i - 1);
}
// Return answer if remaining
// sum can be obtained in suffix
if (x - r >= 0 && x - r <= 9 * k)
{
// Find suffix of length k
// having sum of digits x-r
String suf = helper(k, x - r);
String pre = "";
if (i > 0)
pre = y.substring(0, i);
// Append current character
char cur = (char)(j + (int)'0');
pre += cur;
// Return the result
return pre + suf;
}
}
}
}
// Driver code
public static void main(String[] arg)
{
// Given number and sum
int x = 18;
int y = 99;
// Function call
System.out.print(findMin(x, y));
}
}
// This code is contributed by pratham76
# Python3 program for the
# above approach
# Function to return the
# minimum string of length
# d having the sum of digits s
def helper(d, s):
# Return a string of
# length d
ans = ['0'] * d
for i in range(d - 1,
-1, -1):
# Greedily put 9's
# in the end
if (s >= 9):
ans[i] = '9'
s -= 9
# Put remaining sum
else:
c = chr(s +
ord('0'))
ans[i] = c;
s = 0;
return ''.join(ans);
# Function to find the
# smallest number greater
# than Y whose sum of
# digits is X
def findMin(x, Y):
# Convert number y
# to string
y = str(Y);
n = len(y)
p = [0] * n
# Maintain prefix sum
# of digits
for i in range(n):
p[i] = (ord(y[i]) -
ord('0'))
if (i > 0):
p[i] += p[i - 1];
# Iterate over Y from the
# back where k is current
# length of suffix
n - 1
k = 0
while True:
# Stores current digit
d = 0;
if (i >= 0):
d = (ord(y[i]) -
ord('0'))
# Increase current
# digit
for j in range(d + 1,
10):
# Sum upto current
# prefix
r = ((i > 0) *
p[i - 1] + j);
# Return answer if
# remaining sum can
# be obtained in suffix
if (x - r >= 0 and
x - r <= 9 * k):
# Find suffix of length
# k having sum of digits
# x-r
suf = helper(k,
x - r);
pre = "";
if (i > 0):
pre = y[0 : i]
# Append current
# character
cur = chr(j +
ord('0'))
pre += cur;
# Return the result
return pre + suf;
i -= 1
k += 1
# Driver Code
if __name__ == "__main__":
# Given Number and Sum
x = 18;
y = 99;
# Function Call
print ( findMin(x, y))
# This code is contributed by Chitranayal
// C# program for the above approach
using System;
using System.Text;
using System.Collections;
class GFG{
// Function to return the minimum string
// of length d having the sum of digits s
static string helper(int d, int s)
{
// Return a string of length d
StringBuilder ans = new StringBuilder();
for(int i = 0; i < d; i++)
{
ans.Append("0");
}
for(int i = d - 1; i >= 0; i--)
{
// Greedily put 9's in the end
if (s >= 9)
{
ans[i] = '9';
s -= 9;
}
// Put remaining sum
else
{
char c = (char)(s + (int)'0');
ans[i] = c;
s = 0;
}
}
return ans.ToString();
}
// Function to find the smallest
// number greater than Y
// whose sum of digits is X
static string findMin(int x, int Y)
{
// Convert number y to string
string y = Y.ToString();
int n = y.Length;
ArrayList p = new ArrayList();
for(int i = 0; i < n; i++)
{
p.Add(0);
}
// Maintain prefix sum of digits
for(int i = 0; i < n; i++)
{
p[i] = (int)((int) y[i] - (int)'0');
if (i > 0)
{
p[i] = (int)p[i] +
(int)p[i - 1];
}
}
// Iterate over Y from the back where
// k is current length of suffix
for(int i = n - 1, k = 0;; i--, k++)
{
// Stores current digit
int d = 0;
if (i >= 0)
{
d = (int) y[i] - (int)'0';
}
// Increase current digit
for(int j = d + 1; j <= 9; j++)
{
int r = j;
// Sum upto current prefix
if (i > 0)
{
r += (int) p[i - 1];
}
// Return answer if remaining
// sum can be obtained in suffix
if (x - r >= 0 && x - r <= 9 * k)
{
// Find suffix of length k
// having sum of digits x-r
string suf = helper(k, x - r);
string pre = "";
if (i > 0)
pre = y.Substring(0, i);
// Append current character
char cur = (char)(j + (int)'0');
pre += cur;
// Return the result
return pre + suf;
}
}
}
}
// Driver code
public static void Main(string[] arg)
{
// Given number and sum
int x = 18;
int y = 99;
// Function call
Console.Write(findMin(x, y));
}
}
// This code is contributed by rutvik_56
<script>
// Javascript program for the above approach
// Function to return the minimum String
// of length d having the sum of digits s
function helper(d, s)
{
// Return a String of length d
let ans = [];
for(let i = 0; i < d; i++)
{
ans.push("0");
}
for(let i = d - 1; i >= 0; i--)
{
// Greedily put 9's in the end
if (s >= 9)
{
ans[i] ='9';
s -= 9;
}
// Put remaining sum
else
{
let c = String.fromCharCode(
s + '0'.charCodeAt(0));
ans[i] = c;
s = 0;
}
}
return ans.join("");
}
// Function to find the smallest
// number greater than Y
// whose sum of digits is X
function findMin(x, Y)
{
// Convert number y to String
let y = Y.toString();
let n = y.length;
let p = [];
for(let i = 0; i < n; i++)
{
p.push(0);
}
// Maintain prefix sum of digits
for(let i = 0; i < n; i++)
{
p[i] = y[i].charCodeAt(0) -
'0'.charCodeAt(0);
if (i > 0)
{
p[i] = p[i] + p[i - 1];
}
}
// Iterate over Y from the back where
// k is current length of suffix
for(let i = n - 1, k = 0;; i--, k++)
{
// Stores current digit
let d = 0;
if (i >= 0)
{
d = y[i].charCodeAt(0) -
'0'.charCodeAt(0);
}
// Increase current digit
for(let j = d + 1; j <= 9; j++)
{
let r = j;
// Sum upto current prefix
if (i > 0)
{
r += p[i - 1];
}
// Return answer if remaining
// sum can be obtained in suffix
if (x - r >= 0 && x - r <= 9 * k)
{
// Find suffix of length k
// having sum of digits x-r
let suf = helper(k, x - r);
let pre = "";
if (i > 0)
pre = y.substring(0, i);
// Append current character
let cur = String.fromCharCode(
j + '0'.charCodeAt(0));
pre += cur;
// Return the result
return pre + suf;
}
}
}
}
// Driver code
// Given number and sum
let x = 18;
let y = 99;
// Function call
document.write(findMin(x, y));
// This code is contributed by avanitrachhadiya2155
</script>
Time Complexity: O(log10Y)
Auxiliary Space: O(log10Y)
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