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Java-based LeetCode algorithm problem solutions, regularly updated

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package g1001_1100.s1089_duplicate_zeros;

// #Easy #Array #Two_Pointers #2022_02_24_Time_2_ms_(67.91%)_Space_46.1_MB_(15.10%)

/**
 * 1089 - Duplicate Zeros\.
 *
 * Easy
 *
 * Given a fixed-length integer array `arr`, duplicate each occurrence of zero, shifting the remaining elements to the right.
 *
 * **Note** that elements beyond the length of the original array are not written. Do the above modifications to the input array in place and do not return anything.
 *
 * **Example 1:**
 *
 * **Input:** arr = [1,0,2,3,0,4,5,0]
 *
 * **Output:** [1,0,0,2,3,0,0,4]
 *
 * **Explanation:** After calling your function, the input array is modified to: [1,0,0,2,3,0,0,4]
 *
 * **Example 2:**
 *
 * **Input:** arr = [1,2,3]
 *
 * **Output:** [1,2,3]
 *
 * **Explanation:** After calling your function, the input array is modified to: [1,2,3]
 *
 * **Constraints:**
 *
 * *   1 <= arr.length <= 10⁴
 * *   `0 <= arr[i] <= 9`
**/
public class Solution {
    public void duplicateZeros(int[] arr) {
        int countZero = 0;
        for (int k : arr) {
            if (k == 0) {
                countZero++;
            }
        }
        int len = arr.length + countZero;
        // We just need O(1) space if we scan from back
        // i point to the original array, j point to the new location
        int i = arr.length - 1;
        int j = len - 1;
        while (i < j) {
            // copy twice when hit '0'
            if (arr[i] == 0) {
                if (j < arr.length) {
                    arr[j] = arr[i];
                }
                j--;
            }
            if (j < arr.length) {
                arr[j] = arr[i];
            }
            i--;
            j--;
        }
    }
}