# Next Smaller Element

• Difficulty Level : Medium
• Last Updated : 15 Jul, 2022

Given an array, print the Next Smaller Element (NSE) for every element. The NSE for an element x is the first smaller element on the right side of x in array. Elements for which no smaller element exist (on right side), consider NSE as -1.
Examples:
a) For any array, rightmost element always has NSE as -1.
b) For an array which is sorted in increasing order, all elements have NSE as -1.
c) For the input array [4, 8, 5, 2, 25}, the NSE for each element are as follows.

Element         NSE
4      -->    2
8      -->    5
5      -->    2
2      -->   -1
25     -->   -1

d) For the input array [13, 7, 6, 12}, the next smaller elements for each element are as follows.

  Element        NSE
13      -->    7
7       -->    6
6       -->   -1
12      -->   -1

Method 1 (Simple)
Use two loops: The outer loop picks all the elements one by one. The inner loop looks for the first smaller element for the element picked by outer loop. If a smaller element is found then that element is printed as next, otherwise, -1 is printed.
Thanks to Sachin for providing following code.

## C++

 // Simple C++ program to print  // next smaller elements in a given array #include "bits/stdc++.h" using namespace std;   /* prints element and NSE pair  for all elements of arr[] of size n */ void printNSE(int arr[], int n) {     int next, i, j;     for (i = 0; i < n; i++)     {         next = -1;         for (j = i + 1; j < n; j++)         {             if (arr[i] > arr[j])             {                 next = arr[j];                 break;             }         }         cout << arr[i] << " -- "              << next << endl;     } }   // Driver Code int main() {     int arr[]= {11, 13, 21, 3};     int n = sizeof(arr) / sizeof(arr);     printNSE(arr, n);     return 0; }   // This code is contributed by shivanisinghss2110

## C

 // Simple C program to print next smaller elements // in a given array #include   /* prints element and NSE pair for all elements of arr[] of size n */ void printNSE(int arr[], int n) {     int next, i, j;     for (i=0; i arr[j])             {                 next = arr[j];                 break;             }         }         printf("%d -- %d\n", arr[i], next);     } }   int main() {     int arr[]= {11, 13, 21, 3};     int n = sizeof(arr)/sizeof(arr);     printNSE(arr, n);     return 0; }

## Java

 // Simple Java program to print next // smaller elements in a given array   class Main {     /* prints element and NSE pair for       all elements of arr[] of size n */     static void printNSE(int arr[], int n)     {         int next, i, j;         for (i = 0; i < n; i++) {             next = -1;             for (j = i + 1; j < n; j++) {                 if (arr[i] > arr[j]) {                     next = arr[j];                     break;                 }             }             System.out.println(arr[i] + " -- " + next);         }     }       public static void main(String args[])     {         int arr[] = { 11, 13, 21, 3 };         int n = arr.length;         printNSE(arr, n);     } }

## Python

 # Function to print element and NSE pair for all elements of list def printNSE(arr):       for i in range(0, len(arr), 1):           next = -1         for j in range(i + 1, len(arr), 1):             if arr[i] > arr[j]:                 next = arr[j]                 break                       print(str(arr[i]) + " -- " + str(next))   # Driver program to test above function arr = [11, 13, 21, 3] printNSE(arr)   # This code is contributed by Sunny Karira

## C#

 // Simple C# program to print next // smaller elements in a given array using System;   class GFG {       /* prints element and NSE pair for      all elements of arr[] of size n */     static void printNSE(int[] arr, int n)     {         int next, i, j;         for (i = 0; i < n; i++) {             next = -1;             for (j = i + 1; j < n; j++) {                 if (arr[i] > arr[j]) {                     next = arr[j];                     break;                 }             }             Console.WriteLine(arr[i] + " -- " + next);         }     }       // driver code     public static void Main()     {         int[] arr = { 11, 13, 21, 3 };         int n = arr.Length;           printNSE(arr, n);     } }   // This code is contributed by Sam007

## PHP

  $arr[$j])             {                 $next = $arr[$j];  break;  }  }  echo $arr[$i]." -- ". $next."\n";               } }       // Driver Code     $arr= array(11, 13, 21, 3);  $n = count($arr);  printNSE($arr, \$n);       // This code is contributed by Sam007 ?>

## Javascript

 

Output

11 -- 3
13 -- 3
21 -- 3
3 -- -1

Time Complexity The worst case occurs when all elements are sorted in decreasing order.

Auxiliary Space: O(1)

As constant extra space is used

Method 2 (Using Segment Tree and Binary Search)

This method is also pretty simple if one knows Segment trees and Binary Search. Lets consider an array and lets suppose NSE for is , we simply need to binary search for in range to  will be the first index , such that range minimum of elements from index to ( ) is lesser than .

## C++

 #include  using namespace std;   // Program to find next smaller element for all elements in // an array, using segment tree and binary search   // --------Segment Tree Starts Here-----------------   vector<int> seg_tree;   // combine function for combining two nodes of the tree, in // this case we need to take min of two int combine(int a, int b) { return min(a, b); }   // build function, builds seg_tree based on vector parameter // arr void build(vector<int>& arr, int node, int tl, int tr) {     // if current range consists only of one element, then     // node should be this element     if (tl == tr) {         seg_tree[node] = arr[tl];     }     else {         // divide the build operations into two parts         int tm = (tr - tl) / 2 + tl;           build(arr, 2 * node, tl, tm);         build(arr, 2 * node + 1, tm + 1, tr);           // combine the results from two parts, and store it         // into current node         seg_tree[node] = combine(seg_tree[2 * node],                                  seg_tree[2 * node + 1]);     } }   // query function, returns minimum in the range [l, r] int query(int node, int tl, int tr, int l, int r) {     // if range is invalid, then return infinity     if (l > r) {         return INT32_MAX;     }       // if range completely aligns with a segment tree node,     // then value of this node should be returned     if (l == tl && r == tr) {         return seg_tree[node];     }       // else divide the query into two parts     int tm = (tr - tl) / 2 + tl;       int q1 = query(2 * node, tl, tm, l, min(r, tm));     int q2 = query(2 * node + 1, tm + 1, tr, max(l, tm + 1),                    r);       // and combine the results from the two parts and return     // it     return combine(q1, q2); }   // --------Segment Tree Ends Here-----------------   void printNSE(vector<int> arr, int n) {     seg_tree = vector<int>(4 * n);       // build segment tree initially     build(arr, 1, 0, n - 1);       int q, l, r, mid, ans;     for (int i = 0; i < n; i++) {         // binary search for ans in range [i + 1, n - 1],         // initially ans is -1 representing there is no NSE         // for this element         l = i + 1;         r = n - 1;         ans = -1;           while (l <= r) {             mid = (r - l) / 2 + l;             // q is the minimum element in range [l, mid]             q = query(1, 0, n - 1, l, mid);               // if the minimum element in range [l, mid] is             // less than arr[i], then mid can be answer, we             // mark it, and look for a better answer in left             // half. Else if q is greater than arr[i], mid             // can't be an answer, we should search in right             // half               if (q < arr[i]) {                 ans = arr[mid];                 r = mid - 1;             }             else {                 l = mid + 1;             }         }           // print NSE for arr[i]         cout << arr[i] << " ---> " << ans << "\n";     } }   // Driver program to test above functions int main() {     vector<int> arr = { 11, 13, 21, 3 };     printNSE(arr, 4);     return 0; }

Output

11 ---> 3
13 ---> 3
21 ---> 3
3 ---> -1

Time Complexity : For each of array elements we do a binary search, which includes steps, and each step costs operations [range minimum queries].

Auxiliary Space: O(N)

As extra space is used for storing the elements of the segment tree.

Method 3 (Using Segment Tree and Coordinate Compression)

In this approach, we build a segment tree on indices of compressed array elements:

1. Somewhere along the lines, we would build a array such-that is the smallest index at which is present in input array.
2. Its easy to see that we need to compress the input array so as to build this array because if exceeds (memory limit of online judge) chances are we would get a segmentation fault.
3. To compress we sort the input array, and then for each new value seen in array we map it to a corresponding smaller value, if possible. Use these mapped values to generate a array with same order as input array.
4. So now that we are done with compression, we can begin with the query part:
• Suppose in previous step, we compressed the array to distinct values. Initially set , this signifies no value is processed at any index as of now.
• Traverse the compressed array in reverse order, this would imply that in past we would have only processed elements that are on the right side.
• For , query (and store in ) the smallest index of values using segment tree, this must be the NSE for !
• Update the index of to .
5. We stored the index of NSEs for all array elements, we can easily print NSEs themselves as shown in code.

Note: In implementation we use INT32_MAX instead of -1 because storing INT32_MAX doesn’t affect our min-segment tree and still serves the purpose of identifying unprocessed values.

As extra space is used for storing the elements of the segment tree.

## C++

 #include  using namespace std;   // Program to find next smaller element for all elements in // an array, using segment tree and coordinate compression   // --------Segment Tree Starts Here-----------------   vector<int> seg_tree;   // combine function for combining two nodes of the tree, in // this case we need to take min of two int combine(int a, int b) { return min(a, b); }   // build function, builds seg_tree based on vector parameter // arr void build(vector<int>& arr, int node, int tl, int tr) {     // if current range consists only of one element, then     // node should be this element     if (tl == tr) {         seg_tree[node] = arr[tl];     }     else {         // divide the build operations into two parts         int tm = (tr - tl) / 2 + tl;           build(arr, 2 * node, tl, tm);         build(arr, 2 * node + 1, tm + 1, tr);           // combine the results from two parts, and store it         // into current node         seg_tree[node] = combine(seg_tree[2 * node],                                  seg_tree[2 * node + 1]);     } }   // update function, used to make a point update, update // arr[pos] to new_val and make required changes to segtree void update(int node, int tl, int tr, int pos, int new_val) {     // if current range only contains one point, this must     // be arr[pos], update the corresponding node to new_val     if (tl == tr) {         seg_tree[node] = new_val;     }     else {         // else divide the range into two parts         int tm = (tr - tl) / 2 + tl;           // if pos lies in first half, update this half, else         // update second half         if (pos <= tm) {             update(2 * node, tl, tm, pos, new_val);         }         else {             update(2 * node + 1, tm + 1, tr, pos, new_val);         }           // combine results from both halfs         seg_tree[node] = combine(seg_tree[2 * node],                                  seg_tree[2 * node + 1]);     } }   // query function, returns minimum in the range [l, r] int query(int node, int tl, int tr, int l, int r) {     // if range is invalid, then return infinity     if (l > r) {         return INT32_MAX;     }       // if range completely aligns with a segment tree node,     // then value of this node should be returned     if (l == tl && r == tr) {         return seg_tree[node];     }       // else divide the query into two parts     int tm = (tr - tl) / 2 + tl;       int q1 = query(2 * node, tl, tm, l, min(r, tm));     int q2 = query(2 * node + 1, tm + 1, tr, max(l, tm + 1),                    r);       // and combine the results from the two parts and return     // it     return combine(q1, q2); }   // --------Segment Tree Ends Here-----------------   void printNSE(vector<int> original, int n) {     vector<int> sorted(n);     map<int, int> encode;       // -------Coordinate Compression Starts Here ------       // created a temporary sorted array out of original     for (int i = 0; i < n; i++) {         sorted[i] = original[i];     }     sort(sorted.begin(), sorted.end());       // encode each value to a new value in sorted array     int ctr = 0;     for (int i = 0; i < n; i++) {         if (encode.count(sorted[i]) == 0) {             encode[sorted[i]] = ctr++;         }     }       // use encode to compress original array     vector<int> compressed(n);     for (int i = 0; i < n; i++) {         compressed[i] = encode[original[i]];     }       // -------Coordinate Compression Ends Here ------       // Create an aux array of size ctr, and build a segtree     // based on this array       vector<int> aux(ctr, INT32_MAX);     seg_tree = vector<int>(4 * ctr);       build(aux, 1, 0, ctr - 1);       // For each compressed[i], query for index of NSE and     // update segment tree       vector<int> ans(n);     for (int i = n - 1; i >= 0; i--) {         ans[i] = query(1, 0, ctr - 1, 0, compressed[i] - 1);         update(1, 0, ctr - 1, compressed[i], i);     }       // Print -1 if NSE doesn't exist, otherwise print NSE     // itself       for (int i = 0; i < n; i++) {         cout << original[i] << " ---> ";         if (ans[i] == INT32_MAX) {             cout << -1;         }         else {             cout << original[ans[i]];         }         cout << "\n";     } }   // Driver program to test above functions int main() {     vector<int> arr = { 11, 13, 21, 3 };     printNSE(arr, 4);     return 0; }

Output

11 ---> 3
13 ---> 3
21 ---> 3
3 ---> -1

Time Complexity Auxiliary Space: O(N)

Method 4 (Using Stack)
This problem is similar to next greater element. Here we maintain items in increasing order in the stack (instead of decreasing in next greater element problem).

1. Push the first element to stack.
2. Pick rest of the elements one by one and follow following steps in loop.
• Mark the current element as next.
• If stack is not empty, then compare next with stack top. If next is smaller than top then next is the NSE for the top. Keep popping from the stack while top is greater than next. next becomes the NSE for all such popped elements
• Push next into the stack
3. After the loop in step 2 is over, pop all the elements from stack and print -1 as next element for them.

Note: To achieve same order, we use a stack of pairs, where first element is the value and second element is index of array element.

## C++

 // A Stack based C++ program to find next // smaller element for all array elements #include  using namespace std;   // prints NSE for elements of array arr[] of size n   void printNSE(int arr[], int n) {     stack > s;     vector<int> ans(n);       // iterate for rest of the elements     for (int i = 0; i < n; i++) {         int next = arr[i];           // if stack is empty then this element can't be NSE         // for any other element, so just push it to stack         // so that we can find NSE for it, and continue         if (s.empty()) {             s.push({ next, i });             continue;         }           // while stack is not empty and the top element is         // greater than next         //  a) NSE for top is next, use top's index to         //    maintain original order         //  b) pop the top element from stack           while (!s.empty() && s.top().first > next) {             ans[s.top().second] = next;             s.pop();         }           // push next to stack so that we can find NSE for it           s.push({ next, i });     }       // After iterating over the loop, the remaining elements     // in stack do not have any NSE, so set -1 for them       while (!s.empty()) {         ans[s.top().second] = -1;         s.pop();     }       for (int i = 0; i < n; i++) {         cout << arr[i] << " ---> " << ans[i] << endl;     } }   // Driver program to test above functions int main() {     int arr[] = { 11, 13, 21, 3 };     int n = sizeof(arr) / sizeof(arr);     printNSE(arr, n);     return 0; }

## Java

 // A Stack based Java program to find next // smaller element for all array elements // in same order as input. import java.io.*; import java.lang.*; import java.util.*;   class GFG {     /* prints element and NSE pair for all     elements of arr[] of size n */     public static void printNSE(int arr[], int n)     {         Stack s = new Stack();         HashMap mp             = new HashMap();           /* push the first element to stack */         s.push(arr);           // iterate for rest of the elements         for (int i = 1; i < n; i++) {               if (s.empty()) {                 s.push(arr[i]);                 continue;             }               /* if stack is not empty, then     pop an element from stack.     If the popped element is greater     than next, then     a) print the pair     b) keep popping while elements are     greater and stack is not empty */               while (s.empty() == false                    && s.peek() > arr[i]) {                 mp.put(s.peek(), arr[i]);                 s.pop();             }               /* push next to stack so that we can find             next smaller for it */             s.push(arr[i]);         }           /* After iterating over the loop, the remaining         elements in stack do not have the next smaller         element, so print -1 for them */         while (s.empty() == false) {             mp.put(s.peek(), -1);             s.pop();         }           for (int i = 0; i < n; i++)             System.out.println(arr[i] + " ---> "                                + mp.get(arr[i]));     }       /* Driver program to test above functions */     public static void main(String[] args)     {         int arr[] = { 11, 13, 21, 3 };         int n = arr.length;         printNSE(arr, n);     } }

## Python3

 # A Stack based Python3 program to find next # smaller element for all array elements # in same order as input.using System;   """ prints element and NSE pair for all elements of arr[] of size n """     def printNSE(arr, n):     s = []     mp = {}       # push the first element to stack     s.append(arr)       # iterate for rest of the elements     for i in range(1, n):         if (len(s) == 0):             s.append(arr[i])             continue           """ if stack is not empty, then         pop an element from stack.         If the popped element is greater         than next, then         a) print the pair         b) keep popping while elements are         greater and stack is not empty """         while (len(s) != 0 and s[-1] > arr[i]):             mp[s[-1]] = arr[i]             s.pop()           """ push next to stack so that we can find         next smaller for it """         s.append(arr[i])       """ After iterating over the loop, the remaining     elements in stack do not have the next smaller     element, so print -1 for them """     while (len(s) != 0):         mp[s[-1]] = -1         s.pop()       for i in range(n):         print(arr[i], "--->", mp[arr[i]])     arr = [11, 13, 21, 3] n = len(arr) printNSE(arr, n)   # This code is contributed by decode2207.

## C#

 // A Stack based C# program to find next // smaller element for all array elements // in same order as input.using System; using System; using System.Collections.Generic;   class GFG {     /* prints element and NSE pair for all     elements of arr[] of size n */     public static void printNSE(int[] arr, int n)     {         Stack<int> s = new Stack<int>();         Dictionary<int, int> mp             = new Dictionary<int, int>();           /* push the first element to stack */         s.Push(arr);           // iterate for rest of the elements         for (int i = 1; i < n; i++) {             if (s.Count == 0) {                 s.Push(arr[i]);                 continue;             }             /* if stack is not empty, then             pop an element from stack.             If the popped element is greater             than next, then             a) print the pair             b) keep popping while elements are             greater and stack is not empty */             while (s.Count != 0 && s.Peek() > arr[i]) {                 mp.Add(s.Peek(), arr[i]);                 s.Pop();             }               /* push next to stack so that we can find             next smaller for it */             s.Push(arr[i]);         }           /* After iterating over the loop, the remaining         elements in stack do not have the next smaller         element, so print -1 for them */         while (s.Count != 0) {             mp.Add(s.Peek(), -1);             s.Pop();         }           for (int i = 0; i < n; i++)             Console.WriteLine(arr[i] + " ---> "                               + mp[arr[i]]);     }       // Driver code     public static void Main()     {         int[] arr = { 11, 13, 21, 3 };         int n = arr.Length;         printNSE(arr, n);     } } // This code is contributed by // 29AjayKumar

## Javascript

 

Output

11 ---> 3
13 ---> 3
21 ---> 3
3 ---> -1

Time Complexity: As we use only single for loop and all the elements in the stack are push and popped atmost once.

Auxiliary Space: O(N)

As extra space is used for storing the elements of the stack.

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