Given an integer k and an array op[], in a single operation op[0] will be added to k and then in the second operation k = k + op[1] and so on in a circular manner until k > 0. The task is to print the operation number in which k will be reduced to ? 0. If it impossible to reduce k with the given operations then print -1.
Examples:Â
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Input: op[] = {-60, 10, -100}, k = 100Â
Output: 3Â
Operation 1: 100 – 60 = 40Â
Operation 2: 40 + 10 = 50Â
Operation 3: 50 – 100 = -50
Input: op[] = {1, 1, -1}, k = 10Â
Output: -1
Input: op[] = {-60, 65, -1, 14, -25}, k = 100000Â
Output: 71391Â
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Approach: Count the number of times all the operations can be performed on the number k without actually reducing it to get the result. Then update count = times * n where n is the number of operations. Now, for the remaining operations perform each of the operation one by one and increment count. The first operation when k is reduced to ? 0, print the count.
Below is the implementation of the above approach:Â
Â
C++
// C++ implementation of the approach Â
#include <bits/stdc++.h>using namespace std;Â
Â
int operations(int op[], int n, int k) Â Â Â Â { Â Â Â Â Â Â Â Â int i, count = 0; Â
        // To store the normalized value         // of all the operations         int nVal = 0; Â
        // Minimum possible value for         // a series of operations         int minimum = INT_MAX;         for (i = 0; i < n; i++)         {             nVal += op[i];             minimum = min(minimum , nVal); Â
            // If k can be reduced with             // first (i + 1) operations             if ((k + nVal) <= 0)                 return (i + 1);         } Â
        // Impossible to reduce k         if (nVal >= 0)             return -1; Â
        // Number of times all the operations         // can be performed on k without         // reducing it to <= 0         int times = (k - abs(minimum )) / abs(nVal); Â
        // Perform operations         k = (k - (times * abs(nVal)));         count = (times * n); Â
        // Final check         while (k > 0) {             for (i = 0; i < n; i++) {                 k = k + op[i];                 count++;                 if (k <= 0)                     break;             }         } Â
        return count;     } Â
// Driver codeint main() {Â Â Â Â Â Â Â Â Â Â Â Â Â int op[] = { -60, 65, -1, 14, -25 }; Â Â Â Â Â Â Â Â int n = sizeof(op)/sizeof(op[0]); Â Â Â Â Â Â Â Â int k = 100000; Â
        cout << operations(op, n, k) << endl; }// This code is contributed by Ryuga |
Java
// Java implementation of the approachclass GFG {Â
    static int operations(int op[], int n, int k)    {        int i, count = 0;Â
        // To store the normalized value        // of all the operations        int nVal = 0;Â
        // Minimum possible value for        // a series of operations        int min = Integer.MAX_VALUE;        for (i = 0; i < n; i++) {            nVal += op[i];            min = Math.min(min, nVal);Â
            // If k can be reduced with            // first (i + 1) operations            if ((k + nVal) <= 0)                return (i + 1);        }Â
        // Impossible to reduce k        if (nVal >= 0)            return -1;Â
        // Number of times all the operations        // can be performed on k without        // reducing it to <= 0        int times = (k - Math.abs(min)) / Math.abs(nVal);Â
        // Perform operations        k = (k - (times * Math.abs(nVal)));        count = (times * n);Â
        // Final check        while (k > 0) {            for (i = 0; i < n; i++) {                k = k + op[i];                count++;                if (k <= 0)                    break;            }        }Â
        return count;    }Â
    // Driver code    public static void main(String[] args)    {        int op[] = { -60, 65, -1, 14, -25 };        int n = op.length;        int k = 100000;Â
        System.out.print(operations(op, n, k));    }} |
Python3
# Python3 implementation of the approach def operations(op, n, k):Â
    i, count = 0, 0Â
    # To store the normalized value    # of all the operations    nVal = 0Â
    # Minimum possible value for    # a series of operations    minimum = 10**9    for i in range(n):        nVal += op[i]        minimum = min(minimum , nVal)Â
        # If k can be reduced with        # first (i + 1) operations        if ((k + nVal) <= 0):            return (i + 1)Â
    # Impossible to reduce k    if (nVal >= 0):        return -1Â
    # Number of times all the operations    # can be performed on k without    # reducing it to <= 0    times = (k - abs(minimum )) // abs(nVal)Â
    # Perform operations    k = (k - (times * abs(nVal)))    count = (times * n)Â
    # Final check    while (k > 0):        for i in range(n):            k = k + op[i]            count += 1            if (k <= 0):                breakÂ
    return countÂ
# Driver codeop = [-60, 65, -1, 14, -25]n = len(op)k = 100000Â
print(operations(op, n, k))Â
# This code is contributed # by mohit kumar |
C#
// C# implementation of the approachusing System;Â
class GFG {Â
    static int operations(int []op, int n, int k)    {        int i, count = 0;Â
        // To store the normalized value        // of all the operations        int nVal = 0;Â
        // Minimum possible value for        // a series of operations        int min = int.MaxValue;        for (i = 0; i < n; i++)        {            nVal += op[i];            min = Math.Min(min, nVal);Â
            // If k can be reduced with            // first (i + 1) operations            if ((k + nVal) <= 0)                return (i + 1);        }Â
        // Impossible to reduce k        if (nVal >= 0)            return -1;Â
        // Number of times all the operations        // can be performed on k without        // reducing it to <= 0        int times = (k - Math.Abs(min)) / Math.Abs(nVal);Â
        // Perform operations        k = (k - (times * Math.Abs(nVal)));        count = (times * n);Â
        // Final check        while (k > 0)         {            for (i = 0; i < n; i++)             {                k = k + op[i];                count++;                if (k <= 0)                    break;            }        }Â
        return count;    }Â
    // Driver code    static void Main()    {        int []op = { -60, 65, -1, 14, -25 };        int n = op.Length;        int k = 100000;Â
        Console.WriteLine(operations(op, n, k));    }}Â
// This code is contributed by mits |
PHP
<?php// PHP implementation of the approach function operations($op, $n, $k) { Â Â Â Â $count = 0; Â
    // To store the normalized value     // of all the operations     $nVal = 0; Â
    // Minimum possible value for     // a series of operations     $minimum = PHP_INT_MAX;     for ($i = 0; $i < $n; $i++)     {         $nVal += $op[$i];         $minimum = min($minimum , $nVal); Â
        // If k can be reduced with         // first (i + 1) operations         if (($k + $nVal) <= 0)             return ($i + 1);     } Â
    // Impossible to reduce k     if ($nVal >= 0)         return -1; Â
    // Number of times all the operations     // can be performed on k without     // reducing it to <= 0     $times = round(($k - abs($minimum )) /                          abs($nVal)); Â
    // Perform operations     $k = ($k - ($times * abs($nVal)));     $count = ($times * $n); Â
    // Final check     while ($k > 0)     {         for ($i = 0; $i < $n; $i++)         {             $k = $k + $op[$i];             $count++;             if ($k <= 0)                 break;         }     } Â
    return $count; } Â
// Driver code$op = array(-60, 65, -1, 14, -25 ); $n = sizeof($op); $k = 100000; Â
echo operations($op, $n, $k); Â
// This code is contributed by ihritik?> |
Javascript
<script>// Javascript implementation of the approachÂ
function operations(op,n,k){    let i, count = 0;           // To store the normalized value        // of all the operations        let nVal = 0;           // Minimum possible value for        // a series of operations        let min = Number.MAX_VALUE;        for (i = 0; i < n; i++) {            nVal += op[i];            min = Math.min(min, nVal);               // If k can be reduced with            // first (i + 1) operations            if ((k + nVal) <= 0)                return (i + 1);        }           // Impossible to reduce k        if (nVal >= 0)            return -1;           // Number of times all the operations        // can be performed on k without        // reducing it to <= 0        let times = Math.floor((k - Math.abs(min)) / Math.abs(nVal));           // Perform operations        k = (k - (times * Math.abs(nVal)));        count = (times * n);           // Final check        while (k > 0) {            for (i = 0; i < n; i++) {                k = k + op[i];                count++;                if (k <= 0)                    break;            }        }           return count;}Â
    // Driver code    let op=[-60, 65, -1, 14, -25];    let n = op.length;    let k = 100000;    document.write(operations(op, n, k));     // This code is contributed by unknown2108.</script> |
Output:Â
71391
Â
Time Complexity: O(n*k)
Auxiliary Space: O(1)
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