Given an array nums[] and an integer X, the task is to reduce X to 0 by removing either the leftmost or the rightmost array elements and subtracting its value from X, minimum number of times. If it’s possible to reduce X to 0, print the count of operations required. Otherwise, return -1.
Examples:
Input: nums[] = {3,2,20,1,1,3}, X = 10
Output: 5
Explanation: X (= 10) – 3 – 1 – 1 – 3 – 2 = 0. Therefore, the number of operations required is 5.Input: nums[] = {1, 1, 4, 2, 3}, X = 5
Output: 2
Explanation: X (= 5) – 3 – 2 = 0. Therefore, the number of operations required is 2.
Approach: The given problem can be solved using Two Pointers technique. Follow the steps below to solve the problem.
- Maintain two pointers left and right, pointing to the ends of the left and right subarrays excluded from X.
- Initialize left to consider the entire array, and right to include nothing.
- Therefore, reduce X by the sum of the array.
- Now, iterate until left reaches the front of the array.
- If the sum of the left and the right subarrays exceeds X (i.e. X < 0), shift left by an index to the left and increase X that element.
- If the sum of the left and the right subarrays is less than X (i.e. X > 0), shift right pointer by an index to the left and reduce X by that element.
- If X is found to be 0 at any point, update the minimum number of operations required.
- Print the minimum number of operations required.
- Below is the implementation of the above approach:
C++14
// C++ program to implement // the above approach#include <bits/stdc++.h>using namespace std;// Function to count the minimum// number of operations required// to reduce x to 0static int minOperations(int nums[], int N, int x){ // If sum of the array // is less than x int sum = 0; for(int i = 0; i < x; i++) sum += nums[i]; if (sum < x) return -1; // Stores the count // of operations int ans = INT_MAX; // Two pointers to traverse the array int l = N - 1, r = N; // Reduce x by the sum // of the entire array x -= sum; // Iterate until l reaches // the front of the array while (l >= 0) { // If sum of elements from // the front exceeds x if (x <= 0) { // Shift towards left x += nums[l]; l -= 1; } // If sum exceeds 0 if (x > 0) { // Reduce x by elements // from the right r -= 1; x -= nums[r]; } // If x is reduced to 0 if (x == 0) { // Update the minimum count // of operations required ans = min(ans, (l + 1) + (N - r)); } } if (ans < INT_MAX) return ans; else return -1;}// Driver Codeint main(){ int nums[] = { 1, 1, 4, 2, 3 }; // Size of the array int N = sizeof(nums) / sizeof(nums[0]); int x = 5; cout << minOperations(nums, N, x); return 0;}// This code is contributed by code_hunt |
Java
// Java program to implement // the above approach import java.util.*;class GFG{ // Function to count the minimum // number of operations required // to reduce x to 0 static int minOperations(int nums[], int x) { // If sum of the array // is less than x int sum = 0; for (int i = 0; i < x; i++) sum += nums[i]; if (sum < x) return -1; // Stores the count // of operations int ans = Integer.MAX_VALUE; // Two pointers to traverse the array int l = nums.length - 1, r = nums.length; // Reduce x by the sum // of the entire array x -= sum; // Iterate until l reaches // the front of the array while (l >= 0) { // If sum of elements from // the front exceeds x if (x <= 0) { // Shift towards left x += nums[l]; l -= 1; } // If sum exceeds 0 if (x > 0) { // Reduce x by elements // from the right r -= 1; x -= nums[r]; } // If x is reduced to 0 if (x == 0) { // Update the minimum count // of operations required ans = Math.min(ans, (l + 1) + (nums.length - r)); } } if (ans < Integer.MAX_VALUE) return ans; else return -1; } // Driver Code public static void main(String[] args) { int[] nums = { 1, 1, 4, 2, 3 }; int x = 5; System.out.println(minOperations(nums, x)); }}// This code is contributed by shubhamsingh10 |
Python3
# Python3 Program to implement# the above approachimport math# Function to count the minimum# number of operations required# to reduce x to 0def minOperations(nums, x): # If sum of the array # is less than x if sum(nums) < x: return -1 # Stores the count # of operations ans = math.inf # Two pointers to traverse the array l, r = len(nums)-1, len(nums) # Reduce x by the sum # of the entire array x -= sum(nums) # Iterate until l reaches # the front of the array while l >= 0: # If sum of elements from # the front exceeds x if x <= 0: # Shift towards left x += nums[l] l -= 1 # If sum exceeds 0 if x > 0: # Reduce x by elements # from the right r -= 1 x -= nums[r] # If x is reduced to 0 if x == 0: # Update the minimum count # of operations required ans = min(ans, (l+1) + (len(nums)-r)) return ans if ans < math.inf else -1# Driver Codenums = [1, 1, 4, 2, 3]x = 5print(minOperations(nums, x)) |
C#
// C# Program to implement// the above approachusing System;class GFG { // Function to count the minimum // number of operations required // to reduce x to 0 static int minOperations(int[] nums, int x) { // If sum of the array // is less than x int sum = 0; for (int i = 0; i < x; i++) sum += nums[i]; if (sum < x) return -1; // Stores the count // of operations int ans = Int32.MaxValue; // Two pointers to traverse the array int l = nums.Length - 1, r = nums.Length; // Reduce x by the sum // of the entire array x -= sum; // Iterate until l reaches // the front of the array while (l >= 0) { // If sum of elements from // the front exceeds x if (x <= 0) { // Shift towards left x += nums[l]; l -= 1; } // If sum exceeds 0 if (x > 0) { // Reduce x by elements // from the right r -= 1; x -= nums[r]; } // If x is reduced to 0 if (x == 0) { // Update the minimum count // of operations required ans = Math.Min(ans, (l + 1) + (nums.Length - r)); } } if (ans < Int32.MaxValue) return ans; else return -1; } // Driver Code public static void Main() { int[] nums = { 1, 1, 4, 2, 3 }; int x = 5; Console.Write(minOperations(nums, x)); }}// This code is contributed by ukasp. |
Javascript
<script>// Javascript program to implement // the above approach// Function to count the minimum// number of operations required// to reduce x to 0function minOperations(nums, x){ // If sum of the array // is less than x let sum = 0; for(let i = 0; i < x; i++) sum += nums[i]; if (sum < x) return -1; // Stores the count // of operations let ans = Number.MAX_VALUE; // Two pointers to traverse the array let l = nums.length - 1, r = nums.length; // Reduce x by the sum // of the entire array x -= sum; // Iterate until l reaches // the front of the array while (l >= 0) { // If sum of elements from // the front exceeds x if (x <= 0) { // Shift towards left x += nums[l]; l -= 1; } // If sum exceeds 0 if (x > 0) { // Reduce x by elements // from the right r -= 1; x -= nums[r]; } // If x is reduced to 0 if (x == 0) { // Update the minimum count // of operations required ans = Math.min(ans, (l + 1) + (nums.length - r)); } } if (ans < Number.MAX_VALUE) return ans; else return -1;}// Driver codelet nums = [ 1, 1, 4, 2, 3 ];let x = 5;document.write(minOperations(nums, x));// This code is contributed by target_2 </script> |
PHP
// PHP program to implement<?php// Function to count the minimum numberfunction minOperations($nums, $N, $x) { // sum of the array $sum = 0; for($i = 0; $i < $x; $i++) $sum += $nums[$i]; if ($sum < $x) return -1; // Stores the count of operations $ans = PHP_INT_MAX; // Two pointers to traverse the array $l = $N - 1; $r = $N; // Reduce x by the sum $x -= $sum; // Iterate until l reaches while ($l >= 0) { // sum of elements from if ($x <= 0) { // Shift towards left $x += $nums[$l]; $l -= 1; } // If sum exceeds 0 if ($x > 0) { $r -= 1; $x -= $nums[$r]; } // x is reduced to 0 if ($x == 0) { // updating the minimum count $ans = min($ans, ($l + 1) + ($N - $r)); } } if ($ans < PHP_INT_MAX) return $ans; else return -1;}// Driver Code $nums = array(1, 1, 4, 2, 3); $N = count($nums); $x = 5; echo minOperations($nums, $N, $x);?> |
Output:
2
Time Complexity: O(N)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
