# Store duplicate keys-values pair and sort the key-value pair by key

Given N key-value pairs that contain duplicate keys and values, the task is to store these pairs and sort the pairs by key.

Examples:

Input :
N : 10
Keys : 5 1 4 6 8 0 6 6 5 5
values: 0 1 2 3 4 5 6 7 8 9
Output :
Keys : 0 1 4 5 5 5 6 6 6 8
values: 5 1 2 0 8 9 3 6 7 4
Explanation:
We have given 10 key, value pairs which contain duplicate keys and values.
The key value pair is {(5, 0), (1, 1), (4, 2), (6, 3), (8, 4), (0, 5), (6, 6),
(6, 7), (5, 8), (5, 9)} and we want to store these key value pair and sort these
key value pair by keys. So, the expected output is {(0, 5), (1, 1), (4, 2), (5, 0),
(5, 8), (5, 9), (6, 3), (6, 6), (6, 7), (8, 4)}. Because the sorted
increasing order of keys is {0, 1, 4, 5, 5, 5, 6, 6, 6, 8}.

Approach:

To solve the problem mentioned above we can use separate arrays to store keys and values and then we can simply sort keys by Merge sort algorithm. We can use any sorting Algorithm but Mergesort is the fastest standard sort algorithm and parallelly we can perform the same operations on values array which is performed on the keys so that the key-value pair will stay at the same index on both the array.

Below is the implementation of the above approach:

 `// Java program to Store duplicate ` `// keys-values pair and sort the ` `// key-value pair by key ` `import` `java.util.*; ` `import` `java.lang.*; ` `import` `java.io.*; ` ` `  `class` `Solution { ` ` `  `    ``// Merges two subarrays of arr[]. ` `    ``// First subarray is arr[l..m] ` `    ``// Second subarray is arr[m+1..r] ` `    ``void` `merge(``int` `arrk[], ``int` `l, ``int` `m, ` `            ``int` `r, ``int` `arrv[]) ` `    ``{ ` `        ``// Sizes of two subarrays ` `        ``// that are to be merged ` `        ``int` `n1 = m - l + ``1``; ` `        ``int` `n2 = r - m; ` ` `  `        ``/* Create temporary arrays */` `        ``int` `L[] = ``new` `int``[n1]; ` `        ``int` `R[] = ``new` `int``[n2]; ` `        ``int` `Lk[] = ``new` `int``[n1]; ` `        ``int` `Rk[] = ``new` `int``[n2]; ` ` `  `        ``/* Copy data to temporary arrays */` `        ``for` `(``int` `i = ``0``; i < n1; ++i) { ` `            ``L[i] = arrk[l + i]; ` `            ``Lk[i] = arrv[l + i]; ` `        ``} ` `        ``for` `(``int` `j = ``0``; j < n2; ++j) { ` `            ``R[j] = arrk[m + ``1` `+ j]; ` `            ``Rk[j] = arrv[m + ``1` `+ j]; ` `        ``} ` ` `  `        ``// Initial indexes of ` `        ``// first and second subarrays ` `        ``int` `i = ``0``, j = ``0``; ` ` `  `        ``int` `k = l; ` ` `  `        ``while` `(i < n1 && j < n2) { ` `            ``if` `(L[i] <= R[j]) { ` `                ``arrk[k] = L[i]; ` `                ``arrv[k] = Lk[i]; ` ` `  `                ``i++; ` `            ``} ` `            ``else` `{ ` `                ``arrk[k] = R[j]; ` `                ``arrv[k] = Rk[j]; ` ` `  `                ``j++; ` `            ``} ` ` `  `            ``k++; ` `        ``} ` ` `  `        ``/* Copy remaining elements of L[] if any */` `        ``while` `(i < n1) { ` `            ``arrk[k] = L[i]; ` `            ``arrv[k] = Lk[i]; ` ` `  `            ``i++; ` `            ``k++; ` `        ``} ` ` `  `        ``/* Copy remaining elements of R[] if any */` `        ``while` `(j < n2) { ` `            ``arrk[k] = R[j]; ` `            ``arrv[k] = Rk[j]; ` ` `  `            ``j++; ` `            ``k++; ` `        ``} ` `    ``} ` ` `  `    ``// Function that sorts arr[l..r] using merge() ` `    ``void` `sort(``int` `arrk[], ``int` `l, ``int` `r, ``int` `arrv[]) ` `    ``{ ` `        ``if` `(l < r) { ` `            ``// Find the middle point ` `            ``int` `m = (l + r) / ``2``; ` ` `  `            ``// Sort first and second halves ` `            ``sort(arrk, l, m, arrv); ` `            ``sort(arrk, m + ``1``, r, arrv); ` ` `  `            ``// Merge the sorted halves ` `            ``merge(arrk, l, m, r, arrv); ` `        ``} ` `    ``} ` ` `  `    ``/* Function to print array of size n */` `    ``static` `void` `printArray(``int` `arr[]) ` `    ``{ ` `        ``int` `n = arr.length; ` ` `  `        ``for` `(``int` `i = ``0``; i < n; ++i) ` ` `  `            ``System.out.print(arr[i] + ``" "``); ` ` `  `        ``System.out.println(); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `                    ``throws` `java.lang.Exception ` `    ``{ ` ` `  `        ``// Size of Array ` `        ``int` `n = ``10``; ` ` `  `        ``// array of keys ` `        ``int``[] arrk = { ``5``, ``1``, ``4``, ``6``, ``8``, ``0``,  ` `                                    ``6``, ``6``, ``5``, ``5` `}; ` ` `  `        ``// array of values ` `        ``int``[] arrv = { ``0``, ``1``, ``2``, ``3``, ``4``, ``5``,  ` `                                    ``6``, ``7``, ``8``, ``9` `}; ` ` `  `        ``Solution ob = ``new` `Solution(); ` ` `  `        ``ob.sort(arrk, ``0``, n - ``1``, arrv); ` ` `  `        ``System.out.print(``"Keys: "``); ` `        ``printArray(arrk); ` ` `  `        ``System.out.println(); ` ` `  `        ``System.out.print(``"Values: "``); ` `        ``printArray(arrv); ` `    ``} ` `} `

Output:

```Keys: 0 1 4 5 5 5 6 6 6 8

Values: 5 1 2 0 8 9 3 6 7 4
```

Time Complexity: O(N*logN)

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