Modify given array by reducing each element by its next smaller element
Last Updated :
23 Jul, 2025
Given an array arr[] of length N, the task is to modify the given array by replacing each element of the given array by its next smaller element, if possible. Print the modified array as the required answer.
Examples:
Input: arr[] = {8, 4, 6, 2, 3}
Output: 4 2 4 2 3
Explanation: The operations can be performed as follows:
- For arr[0], arr[1] is the next smaller element.
- For arr[1], arr[3] is the next smaller element.
- For arr[2], arr[3] is the next smaller element.
- For arr[3], no smaller element present after it.
- For arr[4], no smaller element present after it.
Input: arr[] = {1, 2, 3, 4, 5}
Output: 1 2 3 4 5
Naive Approach: The simplest approach is to traverse the array and for each element, traverse the remaining elements after it and check if any smaller element is present or not. If found, reduce that element by the first smaller element obtained.
Time Complexity: O(N2)
Auxiliary Space: O(N)
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print the final array
// after reducing each array element
// by its next smaller element
void printFinalPrices(vector<int>& arr)
{
// Stores the resultant array
vector<int> ans;
// Traverse the array
for (int i = 0; i < arr.size(); i++) {
int flag = 1;
for (int j = i + 1; j < arr.size(); j++) {
// If a smaller element is found
if (arr[j] <= arr[i]) {
// Reduce current element by
// next smaller element
ans.push_back(arr[i] - arr[j]);
flag = 0;
break;
}
}
// If no smaller element is found
if (flag == 1)
ans.push_back(arr[i]);
}
// Print the answer
for (int i = 0; i < ans.size(); i++)
cout << ans[i] << " ";
}
// Driver Code
int main()
{
// Given array
vector<int> arr = { 8, 4, 6, 2, 3 };
// Function Call
printFinalPrices(arr);
return 0;
}
// Java program for the above approach
import java.util.*;
class GFG{
// Function to print the final array
// after reducing each array element
// by its next smaller element
static void printFinalPrices(int[] arr)
{
// Stores the resultant array
ArrayList<Integer> ans = new ArrayList<Integer>();
// Traverse the array
for(int i = 0; i < arr.length; i++)
{
int flag = 1;
for(int j = i + 1; j < arr.length; j++)
{
// If a smaller element is found
if (arr[j] <= arr[i])
{
// Reduce current element by
// next smaller element
ans.add(arr[i] - arr[j]);
flag = 0;
break;
}
}
// If no smaller element is found
if (flag == 1)
ans.add(arr[i]);
}
// Print the answer
for(int i = 0; i < ans.size(); i++)
System.out.print(ans.get(i) + " ");
}
// Driver Code
public static void main(String[] args)
{
// Given array
int[] arr = { 8, 4, 6, 2, 3 };
// Function Call
printFinalPrices(arr);
}
}
// This code is contributed by code_hunt
# Python3 program for the above approach
# Function to print the final array
# after reducing each array element
# by its next smaller element
def printFinalarr(arr):
# Stores resultant array
ans = []
# Traverse the given array
for i in range(len(arr)):
flag = 1
for j in range(i + 1, len(arr)):
# If a smaller element is found
if arr[j] <= arr[i]:
# Reduce current element by
# next smaller element
ans.append(arr[i] - arr[j])
flag = 0
break
if flag:
# If no smaller element is found
ans.append(arr[i])
# Print the final array
for k in range(len(ans)):
print(ans[k], end =' ')
# Driver Code
if __name__ == '__main__':
# Given array
arr = [8, 4, 6, 2, 3]
# Function call
printFinalarr(arr)
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to print the final array
// after reducing each array element
// by its next smaller element
static void printFinalPrices(int[] arr)
{
// Stores the resultant array
List<int> ans = new List<int>();
// Traverse the array
for(int i = 0; i < arr.Length; i++)
{
int flag = 1;
for(int j = i + 1; j < arr.Length; j++)
{
// If a smaller element is found
if (arr[j] <= arr[i])
{
// Reduce current element by
// next smaller element
ans.Add(arr[i] - arr[j]);
flag = 0;
break;
}
}
// If no smaller element is found
if (flag == 1)
ans.Add(arr[i]);
}
// Print the answer
for(int i = 0; i < ans.Count; i++)
Console.Write(ans[i] + " ");
}
// Driver code
static void Main()
{
// Given array
int[] arr = { 8, 4, 6, 2, 3 };
// Function Call
printFinalPrices(arr);
}
}
// This code is contributed by divyeshrabadiya07
<script>
// Js program for the above approach
// Function to print the final array
// after reducing each array element
// by its next smaller element
function printFinalPrices( arr)
{
// Stores the resultant array
let ans = [];
// Traverse the array
for (let i = 0; i < arr.length; i++) {
let flag = 1;
for (let j = i + 1; j < arr.length; j++) {
// If a smaller element is found
if (arr[j] <= arr[i]) {
// Reduce current element by
// next smaller element
ans.push(arr[i] - arr[j]);
flag = 0;
break;
}
}
// If no smaller element is found
if (flag == 1)
ans.push(arr[i]);
}
// Print the answer
for (let i = 0; i < ans.length; i++)
document.write(ans[i], " ");
}
// Driver Code
// Given array
let arr = [ 8, 4, 6, 2, 3 ];
// Function Call
printFinalPrices(arr);
</script>
Time Complexity: O(N^2) ,As we are running two nested loops to traverse the array.
Space Complexity: O(N),As we are storing the resultant array.
Efficient Approach: To optimize the above approach, the idea is to use Stack data structure. Follow the steps below to solve the problem:
- Initialize a Stack and an array ans[] of size N, to store the resultant array.
- Traverse the given array over the indices i = N - 1 to 0.
- If the stack is empty, push the current element arr[i] to the top of the stack.
- Otherwise, if the current element is greater than the element at the top of the stack, push it into the stack and then remove elements from the stack, until the stack becomes empty or an element smaller than or equal to arr[i] is found. After that, if the stack is not empty, set ans[i] = arr[i] - top element of the stack and then remove it from the stack.
- Otherwise, remove the top element from the stack and set ans[i] equal to the top element in the stack and then remove it from the stack.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print the final array
// after reducing each array element
// by its next smaller element
void printFinalPrices(vector<int>& arr)
{
// Initialize stack
stack<int> minStk;
// Array size
int n = arr.size();
// To store the corresponding element
vector<int> reduce(n, 0);
for (int i = n - 1; i >= 0; i--) {
// If stack is not empty
if (!minStk.empty()) {
// If top element is smaller
// than the current element
if (minStk.top() <= arr[i]) {
reduce[i] = minStk.top();
}
else {
// Keep popping until stack is empty
// or top element is greater than
// the current element
while (!minStk.empty()
&& (minStk.top() > arr[i])) {
minStk.pop();
}
// If stack is not empty
if (!minStk.empty()) {
reduce[i] = minStk.top();
}
}
}
// Push current element
minStk.push(arr[i]);
}
// Print the final array
for (int i = 0; i < n; i++)
cout << arr[i] - reduce[i] << " ";
}
// Driver Code
int main()
{
// Given array
vector<int> arr = { 8, 4, 6, 2, 3 };
// Function call
printFinalPrices(arr);
return 0;
}
// Java program for the above approach
import java.util.*;
class GFG{
// Function to print the final array
// after reducing each array element
// by its next smaller element
static void printFinalPrices(int[] arr)
{
// Initialize stack
Stack<Integer> minStk = new Stack<>();
// Array size
int n = arr.length;
// To store the corresponding element
int[] reduce = new int[n];
for(int i = n - 1; i >= 0; i--)
{
// If stack is not empty
if (!minStk.isEmpty())
{
// If top element is smaller
// than the current element
if (minStk.peek() <= arr[i])
{
reduce[i] = minStk.peek();
}
else
{
// Keep popping until stack is empty
// or top element is greater than
// the current element
while (!minStk.isEmpty() &&
(minStk.peek() > arr[i]))
{
minStk.pop();
}
// If stack is not empty
if (!minStk.isEmpty())
{
reduce[i] = minStk.peek();
}
}
}
// Push current element
minStk.add(arr[i]);
}
// Print the final array
for(int i = 0; i < n; i++)
System.out.print(arr[i] - reduce[i] + " ");
}
// Driver Code
public static void main(String[] args)
{
// Given array
int[] arr = { 8, 4, 6, 2, 3 };
// Function call
printFinalPrices(arr);
}
}
// This code is contributed by Rajput-Ji
# Python3 program for the above approach
# Function to print the final array
# after reducing each array element
# by its next smaller element
def printFinalPrices(arr):
# Initialize stack
minStk = []
# To store the corresponding element
reduce = [0] * len(arr)
for i in range(len(arr) - 1, -1, -1):
# If stack is not empty
if minStk:
# If top element is smaller
# than the current element
if minStk[-1] <= arr[i]:
reduce[i] = minStk[-1]
else:
# Keep popping until stack is empty
# or top element is greater than
# the current element
while minStk and minStk[-1] > arr[i]:
minStk.pop()
if minStk:
# Corresponding elements
reduce[i] = minStk[-1]
# Push current element
minStk.append(arr[i])
# Final array
for i in range(len(arr)):
print(arr[i] - reduce[i], end =' ')
# Driver Code
if __name__ == '__main__':
# Given array
arr = [8, 4, 6, 2, 3]
# Function Call
printFinalPrices(arr)
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to print the readonly array
// after reducing each array element
// by its next smaller element
static void printFinalPrices(int[] arr)
{
// Initialize stack
Stack<int> minStk = new Stack<int>();
// Array size
int n = arr.Length;
// To store the corresponding element
int[] reduce = new int[n];
for(int i = n - 1; i >= 0; i--)
{
// If stack is not empty
if (minStk.Count != 0)
{
// If top element is smaller
// than the current element
if (minStk.Peek() <= arr[i])
{
reduce[i] = minStk.Peek();
}
else
{
// Keep popping until stack is empty
// or top element is greater than
// the current element
while (minStk.Count != 0 &&
(minStk.Peek() > arr[i]))
{
minStk.Pop();
}
// If stack is not empty
if (minStk.Count != 0)
{
reduce[i] = minStk.Peek();
}
}
}
// Push current element
minStk.Push(arr[i]);
}
// Print the readonly array
for(int i = 0; i < n; i++)
Console.Write(arr[i] - reduce[i] + " ");
}
// Driver Code
public static void Main(String[] args)
{
// Given array
int[] arr = { 8, 4, 6, 2, 3 };
// Function call
printFinalPrices(arr);
}
}
// This code contributed by shikhasingrajput
<script>
// javascript program for the above approach
// Function to print the final array
// after reducing each array element
// by its next smaller element
function printFinalPrices(arr)
{
// Initialize stack
var minStk = []
// Array size
var n = arr.length;
var i;
// To store the corresponding element
var reduce = Array(n).fill(0);
for (i = n - 1; i >= 0; i--) {
// If stack is not empty
if (minStk.length>0) {
// If top element is smaller
// than the current element
if (minStk[minStk.length-1] <= arr[i]) {
reduce[i] = minStk[minStk.length-1];
}
else {
// Keep popping until stack is empty
// or top element is greater than
// the current element
while (minStk.length>0
&& (minStk[minStk.length-1] > arr[i])) {
minStk.pop();
}
// If stack is not empty
if (minStk.length>0) {
reduce[i] = minStk[minStk.length-1];
}
}
}
// Push current element
minStk.push(arr[i]);
}
// Print the final array
for (i = 0; i < n; i++)
document.write(arr[i] - reduce[i] + " ");
}
// Driver Code
// Given array
var arr = [8, 4, 6, 2, 3];
// Function call
printFinalPrices(arr);
// This code is contributed by ipg2016107.
</script>
Time Complexity: O(N)
Auxiliary Space: O(N)
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