Have implemented the program to find the minimum number of positive deci-binary numbers needed so that they sum up to n.

Author: rajgohil07

.idea/workspace.xml

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+/*
+Partitioning Into Minimum Number Of Deci-Binary Numbers
+Link: https://leetcode.com/problems/partitioning-into-minimum-number-of-deci-binary-numbers/description/
+
+A decimal number is called deci-binary if each of its digits is either 0 or 1 without any leading zeros. For example, 101 and 1100 are deci-binary, while 112 and 3001 are not.
+
+Given a string n that represents a positive decimal integer, return the minimum number of positive deci-binary numbers needed so that they sum up to n.
+
+Example 1:
+Input: n = "32"
+Output: 3
+Explanation: 10 + 11 + 11 = 32
+
+Example 2:
+Input: n = "82734"
+Output: 8
+
+Example 3:
+Input: n = "27346209830709182346"
+Output: 9
+
+Constraints:
+1 <= n.length <= 105
+n consists of only digits.
+n does not contain any leading zeros and represents a positive integer.
+ */
+
+package com.raj;
+
+public class PartitioningIntoMinimumNumberOfDeciBinaryNumbers {
+    public static void main(String[] args) {
+        // Initialization.
+        String n = "32";
+        int ans = 0;
+
+        // Logic.
+        for (char a : n.toCharArray()) {
+            if (ans < a - '0') {
+                ans = a - '0';
+            }
+        }
+
+        // Display the result.
+        System.out.println(ans + " minimum number of positive deci-binary numbers needed so that they sum up to " + n + ".");
+    }
+}